Question

Find the cost of heating $4.6\text{kg}$ of water from ${25}^{\circ }\text{C}$ to the boiling point, assuming that no energy is wasted. Electrical energy costs $25\text{Paisa}$ per $\text{kWh}$.
[Take specific heat capacity of water, $s=1{\text{cal g}}^{-1}{}^{\circ }{\text{C}}^{-1}$]

• A. $5\text{Paisa}$
• B. $10\text{Paisa}$
• C. $15\text{Paisa}$
• D. $50\text{Paisa}$

Answer 1

The correct option is B $10\text{Paisa}$

Given:
$m=4.6\text{kg};s=1{\text{cal g}}^{-1}{}^{\circ }{\text{C}}^{-1}$
$\Delta T=(100-25)={75}^{\circ }\text{C}$

The amount of heat required is,

$H=ms\Delta T$

$H=(4.6\times {10}^3\times 1\times 75)\text{cal}$

Total electrical energy consumed is,

$W_e=(4.2\times 4.6\times {10}^3\times 1\times 75)\text{J}$

$\text{Power}=\frac{W}{t}$

$P=\frac{\left(4.2\right)(4.6\times {10}^3)\left(75\right)}{60\times 60\times {10}^3}\text{kWh}=0.4\text{kWh}$

Total cost is,

$C=0.4\times 25=10\text{paisa}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{B}\right)$ is the correct answer.