Answer :
880.209 kilojoules of heat are required to heat 2550 grams of water from 17.5°C to 100.0°C.
HOW TO CALCULATE SPECIFIC HEAT:
The amount of energy released or absorbed by a substance can be calculated using the following expression:
Q = m × c × ∆T
where;
- Q = quantity of heat in Joules
- m = mass of substance
- ∆T = change in temperature
According to this question;
- m = 2550g
- ∆T = 100 - 17.5 = 82.5°C
Q = 2550 × 4.184 × 82.5
Q = 880,209J
Q = 880.209kJ.
Therefore, 880.209 kilojoules of heat are required to heat 2550 grams of water from 17.5°C to 100.0°C.
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