Question
Question: 10 gm of ice at –20<sup>0</sup>C is added to 10 gm of water at 50<sup>0</sup>C. Specific heat of wat...
10 gm of ice at –200C is added to 10 gm of water at 500C. Specific heat of water = 1 cal/gm-0C, specific heat of ice = 0.5 cal/gm-0C. Latent heat of ice = 80 cal/gm. Then resulting temperature is –
A
– 200C
B
150C
C
00C
D
50C
Answer
00C
Explanation
Solution
Heat gain by (1) = Heat Lost by other.
Q (Tc = 00C)