Chapter 17

Heat is the amount energy transfer due to a temperature difference. All other forms of energy transfer are classified as work. In the picture below, heat is flowing from the hot object to the cold object.

People used to think that heat was not the same as energy. Therefore they invented a separate unit for heat, the calorie. Terminology: cal : little cal Cal : big cal (this is the unit found on food labels)

Joule proved that heat is equivalent to energy by the following experiment: He proved that one can raise the temperature of water by doing mechanical work (which involves no temperature difference or heating). He found it took 4.186J of energy to raise the temperature of 1g of water by 1°C, hence 1cal=4.186J.

Although T K and T C differs by 273 (recall: T K = T C +273 ), when it comes to ΔT the two scales are the same. That is why in c, it does not matter whether you write 1J kg -1 K -1 or 1J kg -1 °C -1. TfTf TiTi ΔT 1°C0°C1°C 20°C10°C 100°C0°C100°C TfTf TiTi ΔT 274K273K1K 293K283K10K 373K273K100K

Make sure you know how to convert among the following units: Specific heat capacity is the amount of heat required to raise the temperature of 1kg of a substance by 1K.

A 80kg man has a fever of 39°C. How much heat is required to raise his temperature by that amount? [Normal temperature: 37°C, c =4190Jkg -1 K -1 ]

First a reminder about moles: Define the molar mass M as the mass of one mole of substance, then n moles of the substances would have the mass of:

Solid Latent heat of fusion Latent heat of vaporization − + There is no ΔT in the above equation, because there is no temperature change during phase change. Latent heat of fusion: Energy required to melt 1kg of solid Latent heat of vaporization: Energy required to vaporize 1kg of liquid

Heat is absorbed during a phase change but the temperature does not change.

To warm ice (specific heat of ice) 2.10kJkg -1 K -1 To melt ice (latent heat of fusion) 334kJkg -1 To warm water (specific heat of water) 4.19kJkg -1 K -1 To vaporize water (latent heat of vaporization) 2260kJkg -1 To warm steam (specific heat of steam) 2.02kJkg -1 K -1

Find the total amount of energy required to heat 3g of water from - 15°C to 155°C. To warm ice 3g×2.10kJkg -1 K -1 ×15K94.5J To melt ice 3g×334kJkg J To warm water 3g×4.19kJkg -1 K -1 ×100K1257J To vaporize water 3g×2260kJkg J To warm steam 3g×2.02kJkg -1 K -1 ×55K333.3J Total 9470J

Which will do a better job cooling your soda, a “cooler” filled with water at 0°C, or a cooler filled with ice at 0°C ? A.Water B.Ice C.The same

When two objects are in thermal contact, the heat lost by one is the heat gain by the other by conservation of energy. For example, if Q H =-150J, then Q C =+150J. In general, we have: If one of the Q is positive, the other must be negative.

When there are more than two objects present, then the conservation law earlier generalizes to:

2kg of water at 300K is mixed with 2kg of water at 350K. Find the final temperature.

1.5kg of X at temperature 400K is dropped into 3kg of water at 300K. The final temperature is 360K. Find the specific heat capacity of X.

A student wants to cool 0.25kg of soda, initially at 25°C, by adding ice initially at -20°C. How much ice should she add so that the final temperature is 0°C with all the ice melted?

The result is part of a more general result called the equipartition of energy, which states that each degree of freedom gives rise to a contribution of ( 1/2)kT per particle. A degree of freedom is each term that appears as x 2 or p 2 in the energy of the particle, or you can think of it as the number of ways each particle can move.

Air at room temperature has f=5.

What happens to U when T is changed? How much U is needed to change T by one unit? (This is the total heat capacity only when there is no work W )