How many calories of heat were released by the cracker?

Question

Rachel burns a 53 gram cracker under a soda can filled with 58.3 grams of water. She took the temperature of the water before she began -- it was 7.8 degrees Celsius. After the cracker was done burning, the temperature of the water was 60.3 degrees Celsius. How many calories of heat were released by the cracker? Round your answer to one digit after the decimal point.

Answer 1

ΔH=mcΔT

ΔH=heat of the reaction, in this case the amount of heat required to change the water temperature

m=mass of water

c=specific heat capacity of water, 1 cal/g°C

ΔT= change in temperature 60.3-7.8= 52.5

ΔH=(58.3)(1)(52.5)= 12793.935 J= 3060.8 calories

Answer 2

q = mC∆T

q = heat = ?

m = mass of water = 58.3 g

C = specific heat of water = 1 cal/gº

∆T = change in temperature = 60.3º - 7.8º = 52.5º

Solve for q:

q = (58.3 g)(1 cal/gº)(52.5º)

q = 3060.8 cal