How many calories of heat were released by the cracker?

Question

Rachel burns a 53 gram cracker under a soda can filled with 78.3 grams of water. She took the temperature of the water before she began -- it was 11.8 degrees Celsius. After the cracker was done burning, the temperature of the water was 67.1 degrees Celsius. How many calories of heat were released by the cracker? Round your answer to one digit after the decimal point.

Answer 1

This question requires you to use the equation q=mcΔT for the specific heat capacity. You use the values given for water in the equation because you are given the change in temperature, mass, and a known constant for the specific heat of water. For ΔT, you subtract the initial temperature from the final temperature to get the change in temperature.

q=mcΔT

q = heat (calories)

m = mass (g)

c = specific heat of water (1 calorie/(g*°C))

ΔT = change in temperature (°C)

-q_{released by cracker} = q_{gained by water}

q _water = (78.3)(1)(67.1-11.8)

q _water = 4330.0 calories

q _cracker = -4330.0 calories