Question

3. A 200-gram aluminum at 20°C placed in 100-gram water at 80°C in a container. The specific heat of aluminum is 0.22 cal/g°C and the specific heat of water is 1 cal/g°C. What is the final temperature of aluminum?​

Answer 1

Explanation:

use Q=ms×(change in temperature)

so we know that heat lost by water is equal to heat gain by aluminum

200×(T-20)×0.22=100×(80-T)×1

44T-880=8000-100T

144T=8880

T=61.67 °C