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Thermal Properties of Matter · NEET Preparation Physics

Calculate the amount of heat required to convert 5text{ g} of ice at -10^circtext{C} into steam at 100^circtext{C}. (Specific heat of ice = 0.5text{…

Practice question from Thermal Properties of Matter (NEET Preparation Physics), with the correct answer and a step-by-step solution.

hard · MCQ · Q-5325EB

Calculate the amount of heat required to convert 5 g5\text{ g} of ice at −10∘C-10^\circ\text{C} into steam at 100∘C100^\circ\text{C}. (Specific heat of ice =0.5 cal/g∘C= 0.5\text{ cal/g}^\circ\text{C}, latent heat of fusion of ice =80 cal/g= 80\text{ cal/g}, specific heat of water =1 cal/g∘C= 1\text{ cal/g}^\circ\text{C}, latent heat of vaporization =540 cal/g= 540\text{ cal/g})

  1. A.3625 calCorrect answer
  2. B.3600 cal
  3. C.3725 cal
  4. D.3500 cal

Solution

Q1Q_1 (ice −10∘C-10^\circ\text{C} to 0∘C0^\circ\text{C}) =mciΔT=5×0.5×10=25 cal= m c_i \Delta T = 5 \times 0.5 \times 10 = 25\text{ cal}. Q2Q_2 (fusion at 0∘C0^\circ\text{C}) =mLf=5×80=400 cal= m L_f = 5 \times 80 = 400\text{ cal}. Q3Q_3 (water 0∘C0^\circ\text{C} to 100∘C100^\circ\text{C}) =mcwΔT=5×1×100=500 cal= m c_w \Delta T = 5 \times 1 \times 100 = 500\text{ cal}. Q4Q_4 (vaporization at 100∘C100^\circ\text{C}) =mLv=5×540=2700 cal= m L_v = 5 \times 540 = 2700\text{ cal}. Total heat =25+400+500+2700=3625 cal= 25 + 400 + 500 + 2700 = 3625\text{ cal}.

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