QuestionProblem 25: (5\% of Assignment Value) The same heat transfer into identical masses of different substances produces different temperature changes, due to differences in the heat capacity of the various materials. \begin{tabular}{|l|c|c|} \hline \multicolumn{1}{|c|}{ Substances } & \multicolumn{2}{c|}{ Specific heat $(c)$} \\ \hline \multicolumn{1}{|c|}{ Solids } & $\mathrm{J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$ & $\mathrm{kcal} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$ \\ \hline Aluminum & 900 & 0.215 \\ \hline Concrete & 840 & 0.20 \\ \hline Copper & 387 & 0.0924 \\ \hline Glass & 840 & 0.20 \\ \hline Human Body $\left(37^{\circ} \mathrm{C}\right)$ & 3500 & 0.83 \\ \hline Iron, steel & 452 & 0.108 \\ \hline \multicolumn{1}{|c|}{ Liquids } & & \\ \hline Water & 4186 & 1.000 \\ \hline Mercury & 139 & 0.0333 \\ \hline \end{tabular}
Otheexpertta.com - Part (a) $\boldsymbol{v}$
Calculate the final temperature, in degrees Celsius, when 1.25 kcal of heat transfers to 1.25 kg of water, originally at $20^{\circ} \mathrm{C}$. $$T_{\mathrm{w}}=21.00^{\circ} \mathrm{C}$$
Correct! - Part (b) $\boldsymbol{V}$.
Calculate the final temperature, in degrees Celsius, when 1.25 kcal of heat transfers to 1.25 kg of concrete, originally at $20^{\circ} \mathrm{C}$. $$T_{\mathrm{c}}=25.00^{\circ} \mathrm{C}$$ ${ }^{7}$ Correct! - Part (c) $\boldsymbol{V}$
Calculate the final temperature, in degress Celsius, when 1.25 kcal of heat transfers to 1.25 kg of the steel, originally at $20^{\circ} \mathrm{C}$. $$T_{\mathrm{s}}=29.30^{\circ} \mathrm{C}$$ $\checkmark$ Correct! Part (d) Calculate the final temperature, in degrees Celsius, when 1.25 kcal of heat transfers to 1.25 kg of the mercury, originally at $20^{\circ} \mathrm{C}$.