Question

A 155 g sample of an unknown substance was heated from 25°c to 40°c. in the process, the substance absorbed 569 calories of energy. what is the specific heat of the substance?

Answer 1

Answer: $0.24cal/g^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= 569 calories

m= mass of substance = 155 g

c = specific heat capacity = ?

Initial temperature = $T_i$ = 25.0°C

Final temperature = $T_f$  = 40.0°C

Change in temperature ,$\Delta T=T_f-T_i=(40-25)^0C=15^0C$

Putting in the values, we get:

$569=155\times c\times 15^0C$

$c=0.24cal/g^0C$

The specific heat of the substance is $0.24cal/g^0C$

Answer 2

Energy= 2381 joules
heat= Mass(kg) *change in temperature(K) * Cp
2381=0.155*(15)*Cp
Cp=1024 J/kg K