Question

2. The temperature of a silver bar rises by 10.0°C when it absorbs 1.23 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of silver.

Answer 1

We know

$$s=\frac{Q}{{m}{\Delta{t}}}$$

Where We have given

$Q=1.23kJ=1.23×10^{3} joule$ .

$m=525g=0.525kg$ .

$\Delta{t}=10.0^°C$

Now putting all values in above equation ,We have

$$s=\frac{1.23×10^3joule}{0.525kg×10^°C}$$

$$s=2.34×10^2J-kg^{-1}-°C^{-1}$$

Thus the value of specific heat is 2.34×10²J-kg^-1-°C^-1.