How To Find Final Temperature In Heat Capacity

Learning Objectives

By the end of this department, you will exist able to:

• Observe heat transfer and change in temperature and mass.
• Calculate final temperature after oestrus transfer between two objects.

One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase alter and that no piece of work is done on or by the arrangement. Experiments prove that the transferred heat depends on three factors—the change in temperature, the mass of the organisation, and the substance and phase of the substance.

Figure 1. The heat Q transferred to cause a temperature change depends on the magnitude of the temperature alter, the mass of the arrangement, and the substance and phase involved. (a) The amount of rut transferred is directly proportional to the temperature alter. To double the temperature alter of a mass thou, you need to add twice the oestrus. (b) The amount of heat transferred is as well directly proportional to the mass. To cause an equivalent temperature modify in a doubled mass, you need to add twice the heat. (c) The corporeality of heat transferred depends on the substance and its phase. If information technology takes an amount Q of oestrus to crusade a temperature change ΔT in a given mass of copper, information technology volition take x.8 times that corporeality of estrus to cause the equivalent temperature alter in the same mass of h2o assuming no phase alter in either substance.

The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the rut is proportional to the mass of the substance and the temperature change. The transferred heat too depends on the substance so that, for example, the estrus necessary to raise the temperature is less for alcohol than for water. For the aforementioned substance, the transferred estrus also depends on the stage (gas, liquid, or solid).

Heat Transfer and Temperature Change

The quantitative relationship between heat transfer and temperature change contains all three factors:Q =mcΔT, where Q is the symbol for rut transfer, k is the mass of the substance, and ΔT is the modify in temperature. The symbol c stands for specific heat and depends on the textile and phase. The specific heat is the corporeality of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific estrus c is a property of the substance; its SI unit is J/(kg ⋅ K) or J/(kg ⋅ ºC). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. If oestrus transfer is measured in kilocalories, so the unit of specific heat is kcal/(kg ⋅ ºC).

Values of specific oestrus must generally be looked up in tables, because at that place is no simple way to summate them. In full general, the specific heat also depends on the temperature. Tabular array 1 lists representative values of specific estrus for various substances. Except for gases, the temperature and book dependence of the specific oestrus of near substances is weak. We see from this table that the specific heat of water is 5 times that of glass and ten times that of iron, which means that it takes v times as much heat to raise the temperature of water the same amount as for drinking glass and ten times every bit much heat to raise the temperature of water equally for iron. In fact, h2o has one of the largest specific heats of any material, which is important for sustaining life on Earth.

Case ane. Calculating the Required Estrus: Heating Water in an Aluminum Pan

A 0.500 kg aluminum pan on a stove is used to estrus 0.250 liters of water from 20.0ºC to lxxx.0ºC. (a) How much estrus is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the h2o?

Strategy

The pan and the water are ever at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for the heat transfer for the given temperature alter and mass of water and aluminum. The specific estrus values for water and aluminum are given in Table 1.

Solution

Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature.

Calculate the temperature divergence:

ΔT = T _f − T _i = 60.0ºC.

Summate the mass of h2o. Because the density of water is chiliad kg/m³, one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is m _westward  = 0.250 kg.

Calculate the estrus transferred to the h2o. Use the specific oestrus of water in Table one:

Q _westward =m _w c _westΔT = (0.250 kg)(4186 J/kgºC)(threescore.0ºC) = 62.8 kJ.

Calculate the heat transferred to the aluminum. Apply the specific heat for aluminum in Tabular array i:

Q _Al =m _Al c _AlΔT= (0.500 kg)(900 J/kgºC)(threescore.0ºC) = 27.0 × x⁴ J = 27.0 kJ.<

Compare the percentage of estrus going into the pan versus that going into the h2o. First, find the total transferred heat:

Q _Full =Q _w + Q _Al= 62.8 kJ + 27.0 kJ = 89.8 kJ.

Thus, the corporeality of oestrus going into heating the pan is

[latex]\frac{27.0\text{ kJ}}{89.viii\text{ kJ}}\times100\%=30.i\%\\[/latex]

and the amount going into heating the water is

[latex]\frac{62.eight\text{ kJ}}{89.eight\text{ kJ}}\times100\%=69.9\%\\[/latex].

Give-and-take

In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the estrus to achieve the given temperature change for the h2o as compared to the aluminum pan.

Case 2. Calculating the Temperature Increase from the Work Washed on a Substance: Truck Brakes Overheat on Downhill Runs

Effigy 2. The smoking brakes on this truck are a visible bear witness of the mechanical equivalent of rut.

Truck brakes used to command speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the restriction material. This conversion prevents the gravitational potential free energy from being converted into kinetic energy of the truck. The trouble is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur likewise fast for sufficient heat to transfer from the brakes to the environment.

Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J/kg ⋅ ºC if the material retains ten% of the energy from a 10,000-kg truck descending 75.0 1000 (in vertical deportation) at a constant speed.

Strategy

If the brakes are not applied, gravitational potential free energy is converted into kinetic free energy. When brakes are applied, gravitational potential energy is converted into internal free energy of the brake material. Nosotros first calculate the gravitational potential energy (Mgh) that the entire truck loses in its descent and then notice the temperature increase produced in the brake material lone.

Solution

1. Calculate the change in gravitational potential free energy every bit the truck goes downhillMgh = (10,000 kg)(9.lxxx thou/due southⁱⁱ)(75.0 thou) = 7.35 × ten^half-dozen J.
2. Calculate the temperature from the heat transferred using Q =Mgh and [latex]\Delta{T}=\frac{Q}{mc}\\[/latex], where m is the mass of the brake fabric. Insert the values m= 100 kg and c= 800 J/kg ⋅ ºC to find [latex]\Delta{T}=\frac{\left(7.35\times10^six\text{ J}\right)}{\left(100\text{ kg}\correct)\left(800\text{ J/kg}^{\circ}\text{C}\right)}=92^{\circ}C\\[/latex].

Give-and-take

This temperature is close to the boiling betoken of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would probable raise the temperature of the brake material in a higher place the boiling point of water, and so this technique is not practical. However, the same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical free energy (battery).

Table 1. Specific Heats[1] of Diverse Substances
Substances	Specific oestrus (c)
Solids	J/kg ⋅ ºC	kcal/kg ⋅ ºC[2]
Aluminum	900	0.215
Asbestos	800	0.19
Physical, granite (average)	840	0.xx
Copper	387	0.0924
Glass	840	0.20
Gilded	129	0.0308
Man body (average at 37 °C)	3500	0.83
Ice (average, −l°C to 0°C)	2090	0.l
Fe, steel	452	0.108
Lead	128	0.0305
Silver	235	0.0562
Wood	1700	0.four
Liquids
Benzene	1740	0.415
Ethanol	2450	0.586
Glycerin	2410	0.576
Mercury	139	0.0333
Water (15.0 °C)	4186	ane.000
Gases [3]
Air (dry out)	721 (1015)	0.172 (0.242)
Ammonia	1670 (2190)	0.399 (0.523)
Carbon dioxide	638 (833)	0.152 (0.199)
Nitrogen	739 (1040)	0.177 (0.248)
Oxygen	651 (913)	0.156 (0.218)
Steam (100°C)	1520 (2020)	0.363 (0.482)

Note that Example ii is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced past a blow torch instead of mechanically.

Instance 3. Computing the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan

Suppose you pour 0.250 kg of 20.0ºC water (most a cup) into a 0.500-kg aluminum pan off the stove with a temperature of 150ºC. Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later?

Strategy

The pan is placed on an insulated pad so that fiddling heat transfer occurs with the surroundings. Originally the pan and water are non in thermal equilibrium: the pan is at a higher temperature than the water. Rut transfer then restores thermal equilibrium once the water and pan are in contact. Because estrus transfer between the pan and water takes place rapidly, the mass of evaporated h2o is negligible and the magnitude of the heat lost by the pan is equal to the heat gained by the water. The exchange of heat stops once a thermal equilibrium betwixt the pan and the h2o is achieved. The heat exchange can be written as |Q _hot|=Q _{common cold}.

Solution

Utilise the equation for estrus transfer Q =mcΔT to express the estrus lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature:Q _hot =m _Al c _Al(T _f − 150ºC).

Express the heat gained by the h2o in terms of the mass of the water, the specific heat of water, the initial temperature of the h2o and the last temperature:Q _cold=k _W c _W(T _f − 20.0ºC).

Note that Q _hot<0 and Q _cold>0 and that they must sum to zero because the heat lost by the hot pan must be the same every bit the heat gained by the cold h2o:

[latex]\begin{array}{lll}Q_{\text{cold}}+Q_{\text{hot}}&=&0\\Q_{\text{cold}}&=&-Q_{\text{hot}}\\m_{\text{West}}c_{\text{Due west}}\left(T_{\text{f}}-20.0^{\circ}\text{C}\right)&=&-m_{\text{Al}}c_{\text{Al}}\left(T_{\text{f}}-150^{\circ}\text{C}\right)\end{assortment}\\[/latex]

This an equation for the unknown final temperature, T _f.

Bring all terms involving T _f on the left hand side and all other terms on the right hand side. Solve for T _f,

[latex]\displaystyle{T_{\text{f}}}=\frac{m_{\text{Al}}c_{\text{Al}}\left(T_{\text{f}}-150^{\circ}\text{C}\correct)+m_{\text{Westward}}c_{\text{W}}\left(T_{\text{f}}-20.0^{\circ}\text{C}\right)}{m_{\text{Al}}c_{\text{Al}}+m_{\text{W}}c_{\text{Westward}}}\\[/latex],

and insert the numerical values:

[latex]\begin{array}{lll}T_{\text{f}}&=&\frac{\left(0.500\text{ kg}\right)\left(900\text{ J/kg}^{\circ}\text{C}\right)\left(150^{\circ}\text{C}\right)+\left(0.250\text{ kg}\correct)\left(4186\text{ J/kg}^{\circ}\text{C}\right)\left(xx.0^{\circ}\text{C}\right)}{\left(0.500\text{ kg}\right)\left(900\text{ J/kg}^{\circ}\text{C}\right)+\left(0.250\text{ kg}\right)\left(4186\text{ J/kg}^{\circ}\text{C}\right)}\\\text{ }&=&\frac{88430\text{ J}}{1496.five\text{ J}/^{\circ}\text{C}}\\\text{ }&=&59.ane^{\circ}\text{C}\end{array}\\[/latex]

Discussion

This is a typical calorimetry problem—two bodies at unlike temperatures are brought in contact with each other and exchange heat until a mutual temperature is reached. Why is the terminal temperature so much closer to xx.0ºC than 150ºC? The reason is that water has a greater specific heat than most common substances and thus undergoes a pocket-sized temperature change for a given heat transfer. A large body of water, such equally a lake, requires a large corporeality of heat to increment its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. Nevertheless, the h2o temperature does alter over longer times (e.g., summertime to winter).

Take-Home Experiment: Temperature Change of Land and Water

What heats faster, land or water?

To written report differences in heat chapters:

• Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is almost 1.half-dozen times that of water, so y'all can achieve approximately equal masses by using l% more water by volume.)
• Heat both (using an oven or a heat lamp) for the same amount of time.
• Record the final temperature of the two masses.
• At present bring both jars to the same temperature by heating for a longer menstruation of time.
• Remove the jars from the estrus source and measure their temperature every 5 minutes for about thirty minutes.

Which sample cools off the fastest? This activity replicates the phenomena responsible for country breezes and body of water breezes.

Cheque Your Understanding

If 25 kJ is necessary to raise the temperature of a block from 25ºC to 30ºC, how much heat is necessary to rut the block from 45ºC to 50ºC?

Solution

The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case.

Section Summary

• The transfer of heat Q that leads to a change ΔT in the temperature of a body with mass g is Q =mcΔT, where c is the specific heat of the material. This human relationship tin can also be considered as the definition of specific heat.

Conceptual Questions

1. What 3 factors impact the oestrus transfer that is necessary to alter an object'due south temperature?
2. The brakes in a car increase in temperature by ΔT when bringing the car to residuum from a speed v. How much greater would ΔT be if the car initially had twice the speed? You may assume the car to stop sufficiently fast and so that no rut transfers out of the brakes.

Bug & Exercises

1. On a hot day, the temperature of an fourscore,000-L pond pool increases by ane.50ºC. What is the net heat transfer during this heating? Ignore whatsoever complications, such as loss of h2o by evaporation.
2. Bear witness that 1 cal/one thousand · ºC =1 kcal/kg · ºC.
3. To sterilize a l.0-g drinking glass baby bottle, we must raise its temperature from 22.0ºC to 95.0ºC. How much rut transfer is required?
4. The same heat transfer into identical masses of different substances produces different temperature changes. Summate the final temperature when one.00 kcal of heat transfers into 1.00 kg of the post-obit, originally at xx.0ºC: (a) water; (b) physical; (c) steel; and (d) mercury.
5. Rubbing your hands together warms them by converting piece of work into thermal energy. If a woman rubs her hands dorsum and forth for a full of 20 rubs, at a distance of vii.l cm per rub, and with an average frictional force of forty.0 North, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, more often than not in the palms and fingers.
6. A 0.250-kg cake of a pure material is heated from xx.0ºC to 65.0ºC by the addition of 4.35 kJ of free energy. Calculate its specific heat and identify the substance of which information technology is about likely equanimous.
7. Suppose identical amounts of oestrus transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?
8. (a) The number of kilocalories in nutrient is adamant by calorimetry techniques in which the food is burned and the corporeality of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the free energy from burning information technology is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a 54.9ºC temperature increase? (b) Compare your reply to labeling information plant on a package of peanuts and comment on whether the values are consistent.
9. Following vigorous exercise, the torso temperature of an 80.0-kg person is 40.0ºC. At what charge per unit in watts must the person transfer thermal energy to reduce the the trunk temperature to 37.0ºC in 30.0 min, assuming the torso continues to produce energy at the charge per unit of 150 W? i watt = 1 joule/second or 1 W = 1 J/s.
10. Fifty-fifty when shut down after a menstruum of normal utilise, a big commercial nuclear reactor transfers thermal energy at the charge per unit of 150 MW past the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails (1 watt = 1 joule/second or ane W = 1 J/s and 1 MW = 1 megawatt). (a) Summate the rate of temperature increase in degrees Celsius per 2nd (ºC/s) if the mass of the reactor cadre is i.60 × ten^v kg and it has an average specific heat of 0.3349 kJ/kg ⋅ ºC. (b) How long would it take to obtain a temperature increase of 2000ºC, which could cause some metals property the radioactive materials to melt? (The initial charge per unit of temperature increment would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, withal, the temperature increase would slow downward considering the 5 × 10^five-kg steel containment vessel would also begin to heat up.)

Effigy three. Radioactive spent-fuel pool at a nuclear power institute. Spent fuel stays hot for a long time. (credit: U.S. Department of Energy)

Glossary

specific heat: the amount of heat necessary to alter the temperature of 1.00 kg of a substance past i.00 ºC

Selected Solutions to Problems & Exercises

i. 5.02 × 10⁸ J

three. 3.07 × 10³ J

5. 0.171ºC

7. x.8

ix. 617 Westward

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Source: https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/

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