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

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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