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

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

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 = (58.3g)(1 cal /gº)(52.5º)

q = 3060.8 cal