Answer to Question #154378 in Classical Mechanics for John smith

Question #154378

Use the table of specific heat capacities to solve this problem. A bar of aluminum is heated to 90.0C and then placed into 10.0 L of water at 20.0 degrees * C . If the mixture cools to 22.4 degrees * C , what is the mass of the aluminum bar? 


1
Expert's answer
2021-01-13T11:38:24-0500

The water will heated, while the aluminum bar will cooled. We can write the heat balance equation:


cwmw(TmixTw)=cAlmAl(TAlTmix).c_wm_w(T_{mix}-T_w)=c_{Al}m_{Al}(T_{Al}-T_{mix}).

From this equation we can find the mass of the aluminum bar:


mAl=cwmw(TmixTw)cAl(TAlTmix),m_{Al}=\dfrac{c_wm_w(T_{mix}-T_w)}{c_{Al}(T_{Al}-T_{mix})},mAl=4180 JkgC10 kg(22.4 C20 C)900 JkgC(90 C22.4 C)=1.65 kg.m_{Al}=\dfrac{4180\ \dfrac{J}{kg \cdot\\ ^{\circ}C}\cdot 10\ kg\cdot(22.4\ ^{\circ}C-20\ ^{\circ}C)}{900\ \dfrac{J}{kg \cdot\\ ^{\circ}C}\cdot(90\ ^{\circ}C-22.4\ ^{\circ}C)}=1.65\ kg.

Answer:

mAl=1.65 kg.m_{Al}=1.65\ kg.


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