Answer:−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
