Question

A 307-g sample of an unknown mineral was heated to 98.7°C and placed into a calorimeter containing 72.4 g of water at 23.6°C. The heat capacity of the calorimeter was 15.7 J/K. The final temperature in the calorimeter was 32.4°C. What is the specific heat capacity of the mineral?

Answer 1

Answer:

The specific heat capacity of the material (Cp) is

Cp= 0,1378 J/gr K

Explanation:

Assuming

1) the calorimeter is completely insulated ( has no heat losses)

$Q_{min}+ Q_{water} +Q_{cal} = Q_{out} =0$

2) the system reaches equilibrium ( the mineral, water and calorimeter has the same final temperature).

3) The specific heat of water is 4,186 J/gK and remains constant

Therefore

$m_{min}c_{min}(T_{equil}- T_{min})+ m_{w}c_{w}(T_{equil}- T_{cal}) +m_{cal}c_{cal}(T_{equil}- T_{cal}) =0\\\\m_{min}c_{min}(T_{equil}- T_{min})+(m_{w}c_{w} +m_{cal}c_{cal})(T_{equil}- T_{cal}) =0\\\\\\m_{min}c_{min}(T_{equil}- T_{min})= - (m_{w}c_{w} +m_{cal}c_{cal})(T_{equil}- T_{cal})\\\\c_{min}= - (m_{w}c_{w} +m_{cal}c_{cal})(T_{equil}- T_{cal})/[m_{min}(T_{equil}- T_{min})]\\\\\\c_{min}= - (72,4gr*4,186 J/gK +15,7 J/K )(32,4 C- 23,6 C)/[307 gr *(32,4 C- 98,7 C]\\ = 0,1378 J/gK$