Question

when two equal sized pieces of the same metal (heat capacity=C) at different temperatures Th (hot piece) and Tc (cold piece) are brought into thermal contact and isolated from its surroundings. the total change in entropy of the system is

Answer 1

when two equal sized pieces of the same metal (heat capacity=C) at different temperatures Th (hot piece) and Tc (cold piece) are brought into thermal contact and isolated from its surroundings. the total change in entropy of the system is

Explanation:

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}}.

Solution

Use the equation for heat transfer Q=\text{mc}\text{Δ}T to express the heat 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}_{\text{hot}}={m}_{\text{Al}}{c}_{\text{Al}}\left({T}_{\text{f}}-\text{150ºC}\right)\text{.}

Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature:

{Q}_{\text{cold}}={m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)\text{.}

Note that {Q}_{\text{hot}}<0 and {Q}_{\text{cold}}>0 and that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water:

\begin{array}{lll}\hfill {Q}_{\text{cold}}\text{+}{Q}_{\text{hot}}& \text{=}& \text{0,}\\ \hfill {Q}_{\text{cold}}& =& {\text{-Q}}_{\text{hot}},\\ {m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)& =& {\mathrm{-m}}_{\mathrm{Al}}{c}_{\mathrm{Al}}\left({T}_{\text{f}}-\text{150ºC.}\right)\end{array}

This an equation for the unknown final temperature, {T}_{\text{f}}

Bring all terms involving {T}_{\text{f}} on the left hand side and all other terms on the right hand side. Solve for {T}_{\text{f}},

{T}_{\text{f}}=\frac{{m}_{\text{Al}}{c}_{\text{Al}}\left(\text{150ºC}\right)+{m}_{W}{c}_{W}\left(\text{20}\text{.0ºC}\right)}{{m}_{\text{Al}}{c}_{\text{Al}}+{m}_{W}{c}_{W}}\text{,}

and insert the numerical values:

\begin{array}{lll}{T}_{\text{f}}& =& \frac{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{150ºC}\right)\text{+}\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)\left(\text{20.0ºC}\right)}{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)+\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)}\\ & =& \frac{\text{88430 J}}{\text{1496.5 J/ºC}}\\ & =& \text{59}\text{.1ºC.}\end{array}

Discussion

This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to \text{20.0ºC} than \text{150ºC}? The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase 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. However, the water temperature does change over longer times (e.g., summer to winter).