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Heat and Thermodynamics question

2021 · 25 Jul · Shift 1 · Q58
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  5. /2021 · 25 Jul · Shift 1 · Q58

Heat and Thermodynamics question

2021 · 25 Jul · Shift 1 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is : JEE Main 2021 (Online) 25th July Morning Shift Physics - Heat and Thermodynamics Question 234 English
  1. A
    83{8 \over 3}38​
  2. B
    38{3 \over 8}83​
  3. C
    34{3 \over 4}43​
  4. D
    43{4 \over 3}34​
View written solutionFree

Correct answer: B

  1. Use the heating relation

For a body heated at a uniform rate,

P=mcdTdtP = mc\frac{dT}{dt}P=mcdtdT​

where:

  • PPP = power supplied,
  • mmm = mass,
  • ccc = specific heat capacity,
  • dTdt\dfrac{dT}{dt}dtdT​ = slope of the temperature-time graph.

So,

dTdt=Pmc\frac{dT}{dt} = \frac{P}{mc}dtdT​=mcP​

Since both bodies have:

  • equal mass,
  • same uniform heating rate,
  • similar conditions,

we get

dTdt∝1c\frac{dT}{dt} \propto \frac{1}{c}dtdT​∝c1​

Thus, specific heat capacity is inversely proportional to the slope of the TTT vs ttt graph.

  1. Read slopes from the graph

From the graph, the straight-line slopes for bodies AAA and BBB are in the ratio

slopeA:slopeB=8:3\text{slope}_A : \text{slope}_B = 8:3slopeA​:slopeB​=8:3

Since

c∝1slope,c \propto \frac{1}{\text{slope}},c∝slope1​,

we have

cAcB=slopeBslopeA=38\frac{c_A}{c_B} = \frac{\text{slope}_B}{\text{slope}_A} = \frac{3}{8}cB​cA​​=slopeA​slopeB​​=83​
  1. Final answer
cAcB=38\boxed{\frac{c_A}{c_B} = \frac{3}{8}}cB​cA​​=83​​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

This matches our derived answer.

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