Question

Calculate the specific heat of a substance when 63j of energy are transferred as heat to an 8.0 g sample to raise it temperature from 314 k to 340 k

Answer 1

The formula for energy or enthalpy is:

E = m Cp (T2 – T1)

where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature

63 J = 8 g * Cp * (340 K – 314 K)

Cp = 0.3 J / g K

Answer 2

Answer : The specific heat of substance is 0.30 J/g.K

Explanation :

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat = 63 J

m = mass of substance = 8.0 g

`C_w` = specific heat of substance = ?

`T_1` = initial temperature = 314 K

`T_2` = final temperature = 340 K

Now put all the given value in the above formula, we get:

`63J=8.0g* c* (340-314)K`

`c=0.30J/g.K`

Therefore, the specific heat of substance is 0.30 J/g.K