Question

A 42.657 g sample of an aqueous NaCl solution at 30.0C was added to a calorimeter whose heat capacity was 7.50 cal/C. Then 43.578 g of water (specific heat--> 1.00 cal/g*C) at 70.0C was added to the calorimeter, and the system allowed to come to equilibrium. If the final temperature was 49.6C, then what is the specific heat (in cal/g*C.) of the NaCl solution?

Answer 1

$heat \space of \space the \space solution = mc \Delta T$

heat lost by hot water = -(heat gained by NaCl solution + heat gained by the calorimeter)

$43.578 \cdot 1.00 \cdot (49.6 - 70.0)= - (42.657 \cdot c \cdot (49.6 - 30.0) + 7.5 \cdot (49.6- 30.0))$

after simple math specific heat of the NaCl solution is:

c=0.904 cal/g*C