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Heat and Thermodynamics question

2019 · 11 Jan · Shift 1 · Q67
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Heat and Thermodynamics question

2019 · 11 Jan · Shift 1 · Q67

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)
  1. A
    100 g
  2. B
    60 g
  3. C
    50 g
  4. D
    40 g
View written solutionFree

Correct answer: D

  1. Heat lost by water

Water mass =50 g=50\,\text{g}=50g, initial temperature 40∘C40^\circ\text{C}40∘C, final temperature 0∘C0^\circ\text{C}0∘C.

So heat given by water is

Qw=mcΔT=50×4.2×40=8400 JQ_w = m c \Delta T = 50 \times 4.2 \times 40 = 8400\,\text{J}Qw​=mcΔT=50×4.2×40=8400J
  1. Let the mass of ice added be mmm g

Initial temperature of ice is −20∘C-20^\circ\text{C}−20∘C.

To raise mmm g of ice from −20∘C-20^\circ\text{C}−20∘C to 0∘C0^\circ\text{C}0∘C, heat needed is

Q1=m×2.1×20=42m JQ_1 = m \times 2.1 \times 20 = 42m\,\text{J}Q1​=m×2.1×20=42mJ
  1. Ice remaining unmelted

At final equilibrium, temperature is 0∘C0^\circ\text{C}0∘C and 20 g20\,\text{g}20g of ice is still unmelted.

This means only (m−20)(m-20)(m−20) g of ice melted.

Heat needed to melt (m−20)(m-20)(m−20) g ice is

Q2=(m−20)×334Q_2 = (m-20) \times 334Q2​=(m−20)×334
  1. Apply conservation of energy

Heat lost by water = heat gained by ice

8400=42m+334(m−20)8400 = 42m + 334(m-20)8400=42m+334(m−20) 8400=42m+334m−66808400 = 42m + 334m - 66808400=42m+334m−6680 8400+6680=376m8400 + 6680 = 376m8400+6680=376m 15080=376m15080 = 376m15080=376m m=15080376≈40.1 gm = \frac{15080}{376} \approx 40.1\,\text{g}m=37615080​≈40.1g
  1. Closest option
m≈40 gm \approx 40\,\text{g}m≈40g

So the correct option is D.

  1. Check options briefly
  • A: 100 g100\,\text{g}100g → far too large.
  • B: 60 g60\,\text{g}60g → too large.
  • C: 50 g50\,\text{g}50g → not close enough.
  • D: 40 g40\,\text{g}40g → matches calculation.

Hence, the amount of ice added was close to 40 g40\,\text{g}40g.

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