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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

Water is boiled in a container whose bottom has a surface area of...

Water is boiled in a container whose bottom has a surface area of 25 cm225 \text{ cm}^2, a thickness of 1.0 mm1.0 \text{ mm}, and a thermal conductivity of 50 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}. After boiling begins, 100 g100 \text{ g} of water is converted into steam per minute during the steady state. Assuming no heat loss to the atmosphere, determine the temperature of the lower surface of the bottom. The latent heat of vaporization of water is 2.26×106 J kg−12.26 \times 10^6 \text{ J kg}^{-1}.

Options

  1. A160 ^{}^{{}^{\circ}}\text{C}
  2. B103 ^{}^{{}^{\circ}}\text{C}
  3. C130 ^{}^{{}^{\circ}}\text{C}
  4. D70 ^{}^{{}^{\circ}}\text{C}

Correct answer

C. 130 ^{}^{{}^{\circ}}\text{C}

Step-by-step solution

The rate of heat transfer through the bottom of the container is given by Fourier's law of heat conduction:

dQdt=KA(T2−T1)d\dfrac{dQ}{dt} = \dfrac{KA(T_2 - T_1)}{d}

The rate of heat utilized in boiling the water is:

dQdt=mLt\dfrac{dQ}{dt} = \dfrac{mL}{t}

Equating the two rates of heat transfer:

KA(T2−T1)d=mLt\dfrac{KA(T_2 - T_1)}{d} = \dfrac{mL}{t}

Substituting the given values, K = 50 \text{ W m}^{-1} ^{{}^{{}^{\circ}}}\textC^{-1}, A=25×10−4 m2A = 25 \times 10^{-4} \text{ m}^2, d=10−3 md = 10^{-3} \text{ m}, m=0.1 kgm = 0.1 \text{ kg}, t=60 st = 60 \text{ s}, L=2.26×106 J kg−1L = 2.26 \times 10^6 \text{ J kg}^{-1}, and T1=100∘CT_1 = 100 ^{{}^{{}^{\circ}}}\text{C}:

50×25×10−4×(T2−100)10−3=0.1×2.26×10660\dfrac{50 \times 25 \times 10^{-4} \times (T_2 - 100)}{10^{-3}} = \dfrac{0.1 \times 2.26 \times 10^6}{60}

125(T2−100)=22600060125(T_2 - 100) = \dfrac{226000}{60}

125(T2−100)=113003125(T_2 - 100) = \dfrac{11300}{3}

T2−100=11300375≈30.13∘CT_2 - 100 = \dfrac{11300}{375} \approx 30.13 ^{{}^{{}^{\circ}}}\text{C}

T2=100+30.13=130.13∘CT_2 = 100 + 30.13 = 130.13 ^{{}^{{}^{\circ}}}\text{C}

The temperature of the lower surface is approximately 130∘C130 ^{{}^{{}^{\circ}}}\text{C}.
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