Question

125.0 g of an unknown substance is heated to 97.0°C. It is then placed in a calorimeter than contains 250g of water with an initial temperature of 20.0°C.

The final temperature reached by the calorimeter is 23.5°C.
What is the specific heat of the unknown substance? The specific heat of water is 4.18 J/ (°C × g)

A) 0.285 J/ (°C × g)
B) 0.398 J/ (°C × g)
C) 0.729 J/ (°C × g)
D) 1.24 J/ (°C × g)

Answer 1

The heat released by the substance in the calorimeter is equal to the heat absorbed by water which results to the decrease and increase in temperature, respectively.
We use m Cp ΔT to balance the heat involved

(m Cp ΔT) subs in calorimeter = (m Cp ΔT) water
125 g * Cp * (97.0-23.5 ) C = 250 g *(4.18 J/C g)* (23.5-20)
Cp = 0.398 J/Cg

Answer is B