1.
An 45.0 g sample of Ni is placed in thermal contact with a 37.35 g piece of Pt initially at
377.0°C. After thermal equilibrium is achieved the final temperature is 402.0°C. What was
the change in temperature of the Ni? The specific heats are C, (Ns))-0.54 J/g C and CPts)0.13 J/g
C O(A)-4. 995 C O(B)-5.545 C O (C)-5345 °C O(D) 25.0 C O(E)-5. 245 C
Solution
Ans A) -4.995 o
C
The heat lost by Ni will be equal to the heat gained by Pt
So here:
Q = mass x specific heat x change in temperature (T)
Heat gained by Pt :
= 37.35 x 0.13 x (402 - 377)
= 121.3875 J
Heat lost by Ni :
121.3875 = 45.0 x 0.54 x T
T = 4.995 o
C