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

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

Explanation :

Formula used :

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

or,

$Q=m\times c\times (T_2-T_1)$

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:

$63J=8.0g\times c\times (340-314)K$

$c=0.30J/g.K$

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

Answer 2

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