Chemistry: A 13.8 g of zinc is heated to 98.8 c in boiling water and then dropped onto a beaker containing 45.0 g of water at 25.o °C .when the water and metal come to thermal equilibrium the temperature is 27.1°C .what is the specific heat capacity

A 13.8 g of zinc is heated to 98.8 c in boiling

Ответ:

PinkDivaGirl02

08.02.2022

Chemistry

$C_{zinc}=0.400\frac{J}{g\°C}$

Explanation:

Hello!

In this case, by considering thermodynamics and the temperature of zinc, we can infer it is hot while water is cold, it means that the heat lost by the zinc is gained by the water and we can write:

$Q_{Zn}=-Q_{water}$

In terms of mass, specific heat and temperatures we can write:

$m_{Zn}C_{Zn}(T_{eq}-T_{Zn})=-m_{water}C_{water}(T_{eq}-T_{water})$

In such a way, by solving for the specific heat capacity of the zinc we write:

$C_{zinc}=\frac{-m_{water}C_{water}(T_{eq}-T_{water})}{m_{Zn}(T_{eq}-T_{Zn})}$

Thus, by plugging in the given values we obtain:

$C_{zinc}=\frac{-45.0g*4.184\frac{J}{g\°C}* (27.1\°C-98.8\°C)}{13.8g*(27.1\°C-98.8\°C)}\\\\C_{zinc}=0.400\frac{J}{g\°C}$

Best regards!

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A 13.8 g of zinc is heated to 98.8 c in boiling water and then dropped onto a beaker containing 45.0 g of water at 25.o °C .when the water and metal come to thermal equilibrium the temperature is 27.1°C .what is the specific heat capacity

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