A 50.00-g sample of metal at 78.0°C is dropped into cold water. If the metal sample cools to 17.0°C and the specific heat of metal is 0.108 cal/g·°C, how much heat is released?

−329.4 calorie

Explanation:Heat released by metal = q=m_mc_m(T_2-T1)q=m

m

c

m

(T

2

−T1)

m_m=50\ g\\c_m=0.108\ cal\ g^{-1}\ \degree C^{-1}\\q=50\times0.108(17-78)=-329.4\ caloriem

m

=50 g

c

m

=0.108 cal g

−1

 °C

−1

q=50×0.108(17−78)=−329.4 calorie

Here minus sign indicate that heat is released

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