Question

A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J kg⁻¹ K⁻¹ and 2100 J kg⁻¹ K⁻¹ respectively. Density of K-oil = 800 kg m⁻³.

Answer 1

Given:
Volume of water, V = 100 cc = 100$\times$10^$-$3 m³
Change in the temperature of the liquid, ∆θ = 5°C
Time, T = 5 min

For water,

$\frac{ms∆\theta }{t}=\frac{KA}{l}(T_1-T_0)$
$$\begin{gathered} \Rightarrow \frac{ms}{t}=\frac{KA}{l}\frac{(T_1-T_0)}{∆\theta } \\ \Rightarrow \frac{V\rho s}{t}=\frac{KA}{l}\frac{(T_1-T_0)}{∆\theta } \\ \Rightarrow \frac{100\times {10}^{-3}\times 1000\times 4200}{5}=\frac{KA}{l}\frac{(313-T_0)}{∆\theta }...\left(\mathrm{i}\right) \end{gathered}$$

For K-oil,

$\frac{ms}{dt}=\frac{\mathrm{KA}}{l}\frac{9(T_1-T_0)}{∆\mathrm{\theta }}$
$$\begin{gathered} \Rightarrow \frac{ms}{t}=\frac{KA}{l}\frac{(T_1-T_0)}{∆\theta } \\ \Rightarrow \frac{V\rho s}{t}=\frac{KA}{l}\frac{(T_1-T_0)}{∆\theta } \\ \Rightarrow \frac{100\times {10}^{-3}\times 800\times 2100}{t}=\frac{KA}{l}\frac{(313-T_0)}{∆\theta }...\left(\mathrm{ii}\right) \end{gathered}$$

From (i) and (ii),

$$\begin{gathered} \frac{100\times {10}^{-3}\times 800\times 2100}{t}=\frac{100\times {10}^{-3}\times 1000\times 4200}{5} \\ \Rightarrow t=\frac{5\times 800\times 2100}{1000\times 4200}=\frac{2000}{1000} \\ \Rightarrow t=2\min \end{gathered}$$