Question

6. How much heat is lost cooling 25 grams of water from 90°C to 45°C?

Answer 1

4,7025 (kJ).

Explanation:

1) according to the condition: m=25 gr.; t₂=90; t₁=45; c=4.18 J/(g°C);

2) the formula of required heat is

Q=c*m*(t₂- t₁);

3) according to the formula of heat:

Q=4.18*25*45=4702.5 (J).