Question

Rachel burns a 53 gram cracker under a soda can filled with 77.8 grams of water. She took the temperature of the water before she began -- it was 9.3 degrees Celsius. After the cracker was done burning, the temperature of the water was 73.8 degrees Celsius. How many calories of heat were released by the cracker? Round your answer to one digit after the decimal point.

Answer 1

In this question, we have a situation in which we have to use the Calorimetry formula, which is how much heat was released or absorbed (in Joules), after we had a change in temperature of a compound. The formula for this Calorimetry question is:

Q = mcΔT

Where:

Q = is energy as Heat

m = mass in grams, 77.8 grams

c = is the specific heat capacity, 4.184 J/g°C this is for water

ΔT = the change in temperature, calculated as Final Temperature - Initial T

73.8 - 9.3 = 64.5°C

It is important to know that the heat released by the cracker is the same heat absorbed by water but with an opposite mathematical sign, so we have to find out the heat absorbed by water:

Q = 77.8 * 4.184 * 64.5

Q = 20995.7 J is the heat absorbed by water

Therefore the heat released by the cracker is - 20,995.7 J