Question

10 gm of ice at -20°C is added to 10 gm of water at 50°C. Specific heat of water =1
cal/gm-°C, specific heat of ice =
0.5 cal/gm-°C. Latent heat of ice = 80 cal/gm. Then, resulting temperature is:
1.-20°C
2. 15°C
3.0°C
4. 50°C
​

Answer 1

Answer:

C. 0°C

Explanation:

q= mc∆T

(0.5)(10)(T+20) +(1)(10)(T-50) +80(Latent heat of ice)=0

final temperature, T = 0