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Heat and Thermodynamics question

2023 · 11 Apr · Shift 1 · Q50
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  5. /2023 · 11 Apr · Shift 1 · Q50

Heat and Thermodynamics question

2023 · 11 Apr · Shift 1 · Q50

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
1 kg1 \mathrm{~kg}1 kg of water at 100∘C100^{\circ} \mathrm{C}100∘C is converted into steam at 100∘C100^{\circ} \mathrm{C}100∘C by boiling at atmospheric pressure. The volume of water changes from 1.00×10−3 m31.00 \times 10^{-3} \mathrm{~m}^{3}1.00×10−3 m3 as a liquid to 1.671 m31.671 \mathrm{~m}^{3}1.671 m3 as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisaiton =2257 kJ/kg=2257 \mathrm{~kJ} / \mathrm{kg}=2257 kJ/kg, Atmospheric pressure = 1×105 Pa)\left.1 \times 10^{5} \mathrm{~Pa}\right)1×105 Pa)
  1. A
    + 2090 kJ
  2. B
    −-− 2426 kJ
  3. C
    + 2476 kJ
  4. D
    −-− 2090 kJ
View written solutionFree

Correct answer: A

  1. Use the first law of thermodynamics

For the boiling process at atmospheric pressure, Q=ΔU+WQ = \Delta U + WQ=ΔU+W so, ΔU=Q−W\Delta U = Q - WΔU=Q−W

Here:

  • Heat supplied for vaporisation: Q=mL=(1)(2257 kJ/kg)=2257 kJQ = mL = (1)(2257\,\text{kJ/kg}) = 2257\,\text{kJ}Q=mL=(1)(2257kJ/kg)=2257kJ
  • Work done by the system at constant pressure: W=PΔV=P(V2−V1)W = P\Delta V = P(V_2 - V_1)W=PΔV=P(V2​−V1​)
  1. Calculate the change in volume

Given: V1=1.00×10−3 m3V_1 = 1.00\times 10^{-3}\,\text{m}^3V1​=1.00×10−3m3 V2=1.671 m3V_2 = 1.671\,\text{m}^3V2​=1.671m3

Thus, ΔV=1.671−0.001=1.670 m3\Delta V = 1.671 - 0.001 = 1.670\,\text{m}^3ΔV=1.671−0.001=1.670m3

  1. Calculate work done

Given atmospheric pressure: P=1×105 PaP = 1\times 10^5\,\text{Pa}P=1×105Pa

So, W=105×1.670=1.67×105 JW = 10^5 \times 1.670 = 1.67\times 10^5\,\text{J}W=105×1.670=1.67×105J W=167 kJW = 167\,\text{kJ}W=167kJ

  1. Find change in internal energy

ΔU=Q−W=2257−167=2090 kJ\Delta U = Q - W = 2257 - 167 = 2090\,\text{kJ}ΔU=Q−W=2257−167=2090kJ

So, ΔU=+2090 kJ\boxed{\Delta U = +2090\,\text{kJ}}ΔU=+2090kJ​

  1. Evaluate options
  • A: +2090+2090+2090 kJ ✅ Correct
  • B: −2426-2426−2426 kJ ❌
  • C: +2476+2476+2476 kJ ❌
  • D: −2090-2090−2090 kJ ❌

Therefore, the correct option is: A\boxed{\text{A}}A​

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