Question

A water heater raises temperature of $1\text{kg}$ of water by $30{}^{\circ }\text{C}$ in $10\text{min}$. How long will it take the same heater to raise $3\text{kg}$ of oil through the same temperature. Given that specific heat of water is twice of oil.

• A. $15\text{min}$
• B. $5\text{min}$
• C. $30\text{min}$
• D. $60\text{min}$

Answer 1

The correct option is A $15\text{min}$

Given, $m_W=1\text{kg};m_O=3\text{kg};\Delta T=30{}^{\circ }\text{C}$
$t_W=10\text{min};S_W=2S_O$

The rate of heat supplied by the heater is,

$P=\frac{Q}{t}$ and $Q=mS\Delta T$

Where, $P=\text{Power of heater}m=\text{mass of liquid}S=\text{Specific heat capacity}\Delta T=\text{Charge in temperature}Q=\text{Heat energy}$

$P=\frac{Q_W}{t_W}=\frac{Q_O}{t_O}$

$\Rightarrow \frac{m_W{S}_W\Delta T}{t_W}=\frac{m_O{S}_O\Delta T}{t_O}$

$\Rightarrow \frac{1\times \left(2S_O\right)\times 30}{10}=\frac{3\times S_O\times 30}{t_O}$

$\Rightarrow t_O=15\text{min}$

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

Why this question ? This question helps you in revising the concept of calorimetry and understanding the relation between energy and power.