Question

Time taken to heat water upto a temperature of $40^{\circ} C$ (from room temperature) is $t_{1}$ and time taken to heat mustard oil (of same mass and at room temperature) upto a temperature of $40^{\circ} C$ is $t_{2}$, then (given mustard oil has smaller heat capacity)

• A. $t_{1}=t_{2}$
• B. $t_{1} >t_{2}$
• C. $t_{2} >t_{1}$
• D. $t_{1}$ and $t_{2}$ both are less than $10 min$

Answer 1

Answer: (B)

The heat capacity of a substance is $S=\frac{\Delta Q}{\Delta T}$.
The heat capacity of mustard oil is less than that of water for same mass. So, same temperature rise $\left(\Delta T=40^{\circ} C \right)$, the quantity of heat $(\Delta Q)$ would be less than that is required by the same amount of water.
Hence, the time taken by water $\left(t_{1}\right)$ to heat upto $40^{\circ} C$ will be higher than that of mustard oil $\left(t_{2}\right)$, i.e. $t_{1}>t_{2}$.