A 4.44 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 24.8°C. When the granite and water come to the same temperature, what will the temperature be?

Question

A  4.44  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  24.8°C. When the granite and water come to the same temperature, what will the temperature be?

DrBob222 · Answer

heat lost by granite + heat gained by water = 0

[mass granite x specific heat granite x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for Tf.

James · Answer

0 = [4440g*0.803*(Tf-98.8)] + [2.00*4.186*(Tf-24.8)] Tf = 98.6C

DrBob222 · Answer

Many thanks for showing your work. It makes it easy to spot the problem. mass 2.0 L H2O is 2,000 g and not 2 g