Question
Question: 50 gm of ice at 0<sup>0</sup>C is added to 200 gm of water initially at 70<sup>0</sup>C in a colorim...
50 gm of ice at 00C is added to 200 gm of water initially at 700C in a colorimeter of unknown water equivalent, let ''W''. The temperature of mixture becomes 400C. When 80 gm of ice at 00C is further added then finally temperature of mixture becomes 100C. As per the given information, the latent heat of fusion of ice should be -
(Specific heat of water = 1 cal/gm0C)
A
70 cal/gm
B
80 cal/gm
C
90 cal/gm
D
100 cal/gm
Answer
90 cal/gm
Explanation
Solution
For the first mixing.
(50 × L) + (50 × 1 × 40) = (200 + W) × 1 × 30
Ž 50L + 2000 = 5000 + 30W
Ž 50L – 30W = 4000 … (1)
For the second mixing.
(80 × L) + (80 × 1 × 10) = (250 + W) × 1 × 30
Ž 80L + 800 = 7500 + 30W
Ž 80L – 30W = 6700 … (2)
From (1) & (2)
30L = 2700
L = 90 Cal/gm