14.2 Temperature change and heat capacity  (Page 3/6)

 Page 3 / 6

Solution

1. Calculate the change in gravitational potential energy as the truck goes downhill
$\text{Mgh}=\left(\text{10,000 kg}\right)\left(9\text{.}\text{80 m}{\text{/s}}^{2}\right)\left(\text{75.0 m}\right)=\text{7.}\text{35}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J.}$
2. Calculate the temperature from the heat transferred using $Q\text{=}\text{Mgh}$ and
$\text{Δ}T=\frac{Q}{\text{mc}}\text{,}$

where $m$ is the mass of the brake material. Insert the values $m=\text{100 kg}$ and $c=\text{800 J/kg}\cdot \text{ºC}$ to find

Discussion

This temperature is close to the boiling point 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 likely raise the temperature of the brake material above the boiling point of water, 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 energy (battery).

Specific heats The values for solids and liquids are at constant volume and at $\text{25ºC}$ , except as noted. Of various substances
Substances Specific heat ( c )
Solids J/kg⋅ºC kcal/kg⋅ºC These values are identical in units of $\text{cal/g}\cdot ºC$ .
Aluminum 900 0.215
Asbestos 800 0.19
Concrete, granite (average) 840 0.20
Copper 387 0.0924
Glass 840 0.20
Gold 129 0.0308
Human body (average at 37 °C) 3500 0.83
Ice (average, -50°C to 0°C) 2090 0.50
Iron, steel 452 0.108
Silver 235 0.0562
Wood 1700 0.4
Liquids
Benzene 1740 0.415
Ethanol 2450 0.586
Glycerin 2410 0.576
Mercury 139 0.0333
Water (15.0 °C) 4186 1.000
Gases ${c}_{\text{v}}$ at constant volume and at $\text{20}\text{.}0ºC$ , except as noted, and at 1.00 atm average pressure. Values in parentheses are ${c}_{\text{p}}$ at a constant pressure of 1.00 atm.
Air (dry) 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 [link] is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically.

Calculating the final temperature when heat is transferred between two bodies: pouring cold water in a hot pan

Suppose you pour 0.250 kg of $\text{20}\text{.0ºC}$ water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of $\text{150ºC}$ . 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 little heat transfer occurs with the surroundings. Originally the pan and water are not in thermal equilibrium: the pan is at a higher temperature than the water. Heat transfer then restores thermal equilibrium once the water and pan are in contact. Because heat transfer between the pan and water takes place rapidly, the mass of evaporated water 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 between the pan and the water is achieved. The heat exchange can be written as $\mid {Q}_{\text{hot}}\mid ={Q}_{\text{cold}}$ .

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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