Question

It takes 11. 2 kj of energy to raise the temperature of 145 g of benzene from 23. 0°c to 68. 0°c. what is the specific heat of benzene?

Answer 1

If we takes 11. 2 kj of energy to raise the temperature of 145 g of benzene from 23. 0°c to 68. 0°c, then the specific heat of benzene is

1716.475 J/kg.°C.

Specific Heat capacity is given as

Q = cm(t₂-t₁) ................$equation 1$

Where Q = quantity of heat, c = specific heat of benzene, m = mass of benzene, t₁  = initial temperature, t₂ = final temperature.

$C$ = $\frac{Q}{m_{(t2-t1)} } ...........equation 2$

Given:

$Q = 11.2 kJ = 11200 J \\\\m= 145\\\\g = 0.145 kg\\\\t_{2} = 68.0 °C\\\\t_{1} = 23.0 °C$

Substitute into equation 2.

$C = \frac{11200}{[0.145(68-23)]}\\\\ C= \frac{11200}{6.525}\\\\ C= 1716.475 J/kg °C$

Hence the specific heat of benzene = $1716.475 J/kg °C$

Learn more about specific heat: brainly.com/question/21041726

#SPJ4