A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays \log _2( P \over P_0 ) , where P0 is a constant. When the metal surface is at a temperature of 487oC, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767oC?

Heat and Thermodynamics PYQ — JEE Advanced 2016 · Shift 1 · Q38 with Solution

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Correct answer: 9

For a surface with constant emissivity,

$P \propto T^4$

where $T$ is the absolute temperature in kelvin.

So,

$\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$

Given:

$T_1 = 487^\circ C = 487 + 273 = 760\,K$ $T_2 = 2767^\circ C = 2767 + 273 = 3040\,K$

Thus,

$\frac{T_2}{T_1} = \frac{3040}{760} = 4$

Hence,

$\frac{P_2}{P_1} = 4^4 = 256 = 2^8$

The sensor displays

$S = \log_2\left(\frac{P}{P_0}\right)$

At $487^\circ C$ , the reading is 1, so

$\log_2\left(\frac{P_1}{P_0}\right)=1$

which means

$\frac{P_1}{P_0}=2$

Now at the higher temperature,

$\frac{P_2}{P_0} = \frac{P_2}{P_1}\cdot \frac{P_1}{P_0} = 2^8 \cdot 2 = 2^9$

Therefore the new sensor reading is

$S_2 = \log_2\left(\frac{P_2}{P_0}\right)=\log_2(2^9)=9$

The value displayed by the sensor is

$\boxed{9}$