A 5.57 kg piece of granite with a specific heat of 0.803 J g-1 °C-1 and a temperature of 98.8 °C is placed?
A 5.57 kg piece of granite with a specific heat of 0.803 J g-1 °C-1 and a temperature of 98.8 °C is placed into 2.00 L of water at 19.0 °C. When the granite and water come to the same temperature, what will the temperature be?
I thought the answer was 58.9 °C but that was marked wrong. Can someone please give me an explanation on how to do this?
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q = m c delta T
q of granite = - [ q of water]
mc delta T = -[ mc delta T]
5570 g x 0.803 J g-1 °C-1 x ( Tf - 98.8) C = - [ 2000 g x 4.18 J/g C x (Tf - 19.0) ]
4472.71T f- 441903 = -8360 Tf + 158840
12832.7 Tf = 600743
Tf = 46.8 deg C
This is what I got !