Question

A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains and equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J kg⁻¹ °C⁻¹.

Answer 1

Given:
Mass of iron = 100 g
Water equivalent of calorimeter = 10 g
Mass of water = 240 gm
Let the temperature of surface be $\theta$ °C.

Specific heat capacity of iron = 470 J kg⁻¹ °C⁻¹

Total heat gained = Total heat lost

$$\begin{gathered} \Rightarrow \frac{100}{1000}\times 470\times \left(\theta -60°\right)=\frac{\left(240+10\right)}{1000}\times 4200\times \left(60-20\right) \\ \Rightarrow 47\theta -47\times 60=25\times 42\times 40 \\ \Rightarrow \theta \mathit{}=\frac{42000+2820}{47}=\frac{44820}{47} \\ =953.61°\mathrm{C} \end{gathered}$$