Question

If a liquid takes $30\text{s}$ in cooling from ${95}^{\circ }\text{C}$ to ${90}^{\circ }\text{C}$ and $70\text{s}$ in cooling from ${55}^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$ then temperature of room is -

• A. ${16.5}^{\circ }\text{C}$
• B. ${22.5}^{\circ }\text{C}$
• C. ${28.5}^{\circ }\text{C}$
• D. ${32.5}^{\circ }\text{C}$

Answer 1

The correct option is B ${22.5}^{\circ }\text{C}$

Step 1, Given data

Liquid takes $30\text{s}$ in cooling from ${95}^{\circ }\text{C}$ to ${90}^{\circ }\text{C}$ and $70\text{s}$ in cooling from ${55}^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$

Step 2, Finding the temperature of the room

Let ${\theta }_0$ be the temperature of the room.

We know,

$\frac{{\theta }_1-{\theta }_2}{t}=K(\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0)$

Case 1
For cooling from ${95}^{\circ }\text{C}$ to ${90}^{\circ }\text{C}$
$\Rightarrow \frac{95-90}{30}=K(\frac{95+90}{2}-{\theta }_0)......\left(i\right)$

Case 2
For cooling from ${55}^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$
$\Rightarrow \frac{55-50}{70}=K(\frac{55+50}{2}-{\theta }_0).......\left(ii\right)$

By dividing equation (i) by (ii),

$\frac{7}{3}=\frac{185-2{\theta }_0}{105-2{\theta }_0}\Rightarrow 735-14{\theta }_0=555-6{\theta }_0$

$\Rightarrow 8{\theta }_0=180\Rightarrow {\theta }_0=\frac{180}{8}={22.5}^{\circ }\text{C}$
Hence, $\left(b\right)$ is the correct option.