Broken calorimeter

Question

You have a slightly broken calorimeter. You combust a 10.0g standard that releases exactly 60 kcal of heat inside of the calorimeter filled with 1050 mL of H2O. You measure the initial temperature of water to be 25.0ºC and then combust the material. The final temperature reads 78.0ºC. How much heat did the water absorb?

How much heat did the calorimeter absorb?

Answer 1

Heat absorbed by the water = q = mC∆T

q = heat = ?

m = mass of water = 1.050 kg (assuming a density of 1 g / ml)

C = specific heat of water = 1 kCal / kgº

∆T = change in temperature = 78º - 25º = 53º

Solving for q ...

q = (1.05 kg)(1 kCal/kgº)(53º) = 55.7 kCal

Heat absorbed by the calorimeter is the difference between the expected heat (60 kCal) and that absorbed by the water.

heat absorbed by calorimeter = 60 kCal - 55.7 kCal = 4.3 kCal (not corrected for sig.figs.)

Also note that if the "slightly broken" calorimeter is broken in such a way that some heat escapes, then we can't really tell how much escaped and how much was absorbed by the calorimeter.

Answer 2

Heat absorbed by water is given by the following equation:

Q = cmΔt where c is the specific heat of water (4.184 kJ/K-kg), m is the mass of water in kg (1.050 kg), and Δt is the increase in temperature in Kelvin degrees( 78.0 - 25.0)K.

Q = 4.184 kJ/kg x 1.050 kg x 53.0K = 232.8 kJ absorbed.

Since one kCal = 4.184 kJ, you can calculate the heat absorbed in kCal as follows:

1 kCal / 4.184 kJ x 232.8 kJ = 55.6 kCal absorbed.