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The calorimeter with an electronic thermometer. The calorimeter is at room temperature: $21,9^o C$
Energy is conserved, there is no exchange of work. The amount of heat lost by the warm water is therefore equal to the amount of heat received by the calorimeter.
$4184 \cdot m_{water}(76,3-71,5)$ $ =$ $ C_{calorimeter} \cdot (71,5-21,9)$;
$4185 \cdot 0,5 \cdot(76,3-71,5)$ $ =$ $ C_{calorimeter}\cdot (71,5-21,9)$;
$C_{calorimeter}$ $ =$ $ 202,5 J/K$