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

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

Answer 2

Answer:
`0.24cal/g^0C`

Step-by-step 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* c* \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* c* 15^0C`

`c=0.24cal/g^0C`

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