Question

Rachel burns a 53 gram cracker under a soda can filled with 81.6 grams of water. She took the temperature of the water before she began -- it was 13.2 degrees Celsius. After the cracker was done burning, the temperature of the water was 68.1 degrees Celsius. How many calories of heat were released by the cracker? Round your answer to one digit after the decimal point.

Answer 1

Answer

Q=4479.8 cal

Procedure

To solve the problem you will need to use the specific heat formula

$Q=mc\Delta T$

Where;

Q=heat energy

m=mass

c=specific heat capacity

ΔT=change in temperature

Assuming that the heat released from the cracker of unknown material is equal to the heat absorbed by the water, then we can use the c and m for water in our calculations.

$c_{water}=4.186\frac{J}{g\degree C}$

Substituting the values in our equation we have

$Q=mc\Delta T=81.6\text{ }g(4.186\frac{J}{g\degree C})(68.1-13.2)\degree C=18752.61\text{ }J$

Finally, transform the J to cal

$18752.61\text{ }J\frac{1\text{ }cal}{4.186\text{ }J}=4479.8\text{ cal}$