50 grams of propane is placed into a bomb calorimeter at 20 deg C. It is found that the heat released is 50 kj. What is the temperature of the system?

The heat capacity of the cariometer is 6.00 kj/C

The final temperature is 28 °C.

There are two heat transfers involved.

heat from combustion of propane + heat gained by water = 0

q_1 + q_2 = 0

q_1 + C_calΔT = 0

q_1 = - 50 kJ

q_2 = 6.00 kJ·°C^ (-1) * ΔT = 6.00 ΔT kJ·°C^ (-1)

q_1 + q_2 = - 50 kJ + 6.00 ΔT kJ·°C^ (-1) = 0

ΔT = 50/[6.00 °C^ (-1) ] = 8.33 °C

T_f = T_i + ΔT = 20 °C + 8.33 °C = 28 °C