Question

Ice at 0⁰C is added to 200gm of water initially at 70⁰C in a vacuum flask. When 50 gm of ice has been added and has all melted, the temperature of flask and contents is 40⁰C, When a further 80 gm of ice is added and has all melted, the temperature of whole becomes 10⁰C. Neglecting heat lost to surroundings the latent heat of fusion of ice is :

• A. 80 cal/gm
• B. 90 cal/gm
• C. 70 cal/gm
• D. 540 cal/gm

Answer 1

Answer: (B)

90 cal/gm

Answer 2

Accoding to principle of calorimetry,

ML_F + Ms DT = (msDT)_water + (msDT)_flask

50L_f + 50 × I × (40 – 0)

= 200 × 1 × (70–40) + W (70 – 40)

or 50L_f + 200 = (200 + W) 30

or 5L_f = 400 + 3W ................(i)

Now the system contains(200 + 50) gm of water at 40⁰C, so when further 80 gm of ice is added.

80L_f + 80 × 1 × (10 – 0)

= 250 × 1 × (40 – 10) + W (40 – 10)

or 80L_f + 800 = (250 + W)30

or 80L_f = 670 + 3W ...............(ii)

Solving equation (i) and (ii),

L_f = 90 cal/gm and W = $\frac{50}{3}$gm