Question

5 x 10 ^ 4 * J of heat are added 35 kg of water initially at 12 degrees C What is the water's final temperature?

Answer 1

1 cal / gm-C = 1000 cal /kgm-C specific heat of water

4.19 joules = 1 C

5.0E4 J / 4.19 J/C = 1.2E4 Calories = ΔQ

ΔQ = K M (T2 - T1) where K is the heat capacity

T2 = 12 + 1.2E4 / (1000 * 35)

T2 = 12 + 1.2E4 / 3.5 E4 = 12.34 C (only .34 deg C)