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

Solution

1. Calculate the change in gravitational potential energy as the truck goes downhill
   $\text{Mgh}=\left(\text{10,000 kg}\right)\left(9\text{.}\text{80 m}{\text{/s}}^2\right)\left(\text{75.0 m}\right)=\text{7.}\text{35}\times {\text{10}}^6\text{J.}$
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

   $\text{Δ}T=\frac{\left(7\text{.35}\times {\text{10}}^6\text{J}\right)}{\left(\text{100 kg}\right)\left(\text{800 J/kgºC}\right)}=\text{92ºC.}$

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).

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 $\mid Q_{\text{hot}}\mid =Q_{\text{cold}}$ .