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Sagot :
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) ................[tex]equation 1[/tex]
Where Q = quantity of heat, c = specific heat of benzene, m = mass of benzene, t₁ = initial temperature, t₂ = final temperature.
[tex]C[/tex] = [tex]\frac{Q}{m_{(t2-t1)} } ...........equation 2[/tex]
Given:
[tex]Q = 11.2 kJ = 11200 J \\\\m= 145\\\\g = 0.145 kg\\\\t_{2} = 68.0 °C\\\\t_{1} = 23.0 °C[/tex]
Substitute into equation 2.
[tex]C = \frac{11200}{[0.145(68-23)]}\\\\ C= \frac{11200}{6.525}\\\\ C= 1716.475 J/kg °C[/tex]
Hence the specific heat of benzene = [tex]1716.475 J/kg °C[/tex]
Learn more about specific heat: https://brainly.com/question/21041726
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