6. How much heat is lost cooling 25 grams of water from 90°C to 45°C?
1 answer:
Answer:
4,7025 (kJ).
Step-by-step 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).
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