Question

Mark is measuring the specific heat of a new type of coolant he first warms 75.0 g of this coolant from 23.5*C to 45.0*C then he puts the warm coolant into a calorimeter and cools it back to 23.5 *C the calorimeter calculations show the sample coolant shed 1370 calories of energy. What is the specific heat of the coolant?

Answer 1

Q= 1370 cal

ΔT = 45.0-23.5=21.5

m=75.0 g

Q = mCΔT

$1370=75.0 \times C \times 21.5 \\ 1370 = 1612.5 \times C \\ C = \frac{1370}{1612.5} = 0.849 \;cal/g \; C$

The specific heat of the coolant is 0.849 cal/gC