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

Explanation:

Step 1:

Given data in the question  

Water volume, $V=100 \mathrm{cc}=100 \times 10^{-3} \mathrm{m}^{3}$

The temperature changes for the liquid, $\Delta \theta=5^{\circ} \mathrm{C}$

T = 5 min

T is time  

Step 2:

In case of  water,

$\frac{m s \Delta \theta}{t}=\frac{K A}{l}\left(T_{1}-T_{0}\right)$

$\frac{m s}{t}=\frac{K A}{l} \frac{\left(T_{1}-T_{0}\right)}{\Delta \theta}$

$\frac{100 \times 10^{-3} \times 1000 \times 4200}{5}=\frac{K A}{l} \frac{\left(313-T_{0}\right)}{\Delta \theta}$    ……… (eq)^n  ( i )

In case of k-oil

$\frac{m s}{t}=\frac{K A}{l} \frac{9\left(T_{1}-T_{0}\right)}{\Delta \theta}$  

$\frac{m s}{t}=\frac{K A}{l} \frac{\left(T_{1}-T_{0}\right)}{\Delta \theta}$

$\frac{V p s}{t}=\frac{K A}{l} \frac{\left(T_{1}-T_{0}\right)}{\Delta \theta}$

$\frac{100 \times 10^{-3} \times 800 \times 2100}{t}=\frac{K A}{l} \frac{313-T_{0}}{\Delta \theta}$          ……..(eq)^n  ( ii )

From equation (i) and equation (ii),

$\frac{100 \times 10^{-3} \times 800 \times 2100}{t}=\frac{100 \times 10^{-3} \times 1000 \times 4200}{5}$

$\frac{100 \times 10^{-3} \times 800 \times 2100}{100 \times 10^{-3} \times 1000 \times 4200}=\frac{t}{5}$

$\frac{800 \times 2100}{1000 \times 4200}=\frac{t}{5}$

$\frac{8 \times 21}{10 \times 42}=\frac{t}{5}$

$\frac{8}{10 \times 2}=\frac{t}{5}$

$\frac{8}{20}=\frac{t}{5}$

$t=\frac{40}{20}$

$t=2 m i n$