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# An 45-0 g sample of Ni is placed in thermal contact with a 37-35 g pie.docx

An 45.0 g sample of Ni is placed in thermal contact with a 37.35 g piece of Pt initially at 377.0Â°C. After thermal equilibrium is achieved the final temperature is 402.0Â°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 Â°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
.

An 45.0 g sample of Ni is placed in thermal contact with a 37.35 g piece of Pt initially at 377.0Â°C. After thermal equilibrium is achieved the final temperature is 402.0Â°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 Â°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
.

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### An 45-0 g sample of Ni is placed in thermal contact with a 37-35 g pie.docx

1. 1. An 45.0 g sample of Ni is placed in thermal contact with a 37.35 g piece of Pt initially at 377.0Â°C. After thermal equilibrium is achieved the final temperature is 402.0Â°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 Â°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