Calculating for the Calorimeter Constant

Question

15.0 mL of distilled water at 8.0*C was mixed in a calorimeter with 25.0 mL of distilled water at 22.0*C. The final temperature of the mixture was 17.5*C. Determine the calorimeter constant. The specific heat of water is 4.184 J/g-*C.

When I solve for it I initially get that the calorimeter lost 130 J. I don't know if this is correct since I believe calorimeters are supposed to absorb heat.

Answer 1

Because the cold water was added to warm water in the calorimeter, the calorimeter is already at 25º, so both the water AND the calorimeter will lose heat to the cold water. Thus, ...

heat gained by cold water = heat lost by warm water + heat lost by calorimeter

heat gained by cold water = q = mC∆T = (15 g)(4.184 J/g)(9.5º) = 596 J

heat lost by warm water + heat lost by calorimeter = (25 g)(4.184 J/gº)(4.5º) + C_cal(4.5º)

596 J = 471 J + 4.5C_cal

C_cal = 27.8 J/º

It is true that the calorimeter absorbs heat, but then it MUST also release heat, which is the case in the current problem. So you probably did the problem correctly but didn't divide by the change in temperature. Calorimeter constants usually have units of J/degree or something similar. If you divide your answer by 4.5 degrees, you get 28.9 J/degree which almost agrees with my answer.