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Calculate the amount of heat transferred when 550 grams of water warms from an initial temperature of 4.0 ºC to a final temperature of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC. Please express you answer as a number in units of kJ, which is already provided after the answer box, and be careful about the sign (is the answer a positive or negative number?) since that shows the direction of heat flow.
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I think the answer should be 550 * 21 * 4.184 but it is marked wrong Why ?
\[Q _{H20}= 4.184 J/g ºC \times 550g \times 21.0 ºC = xxx J/1000= xxx Kj (positive)\]
with that equation you get joules. 1KJ= 10E+3 J
So the answer would be 48325.2 / 1000 = 4.83252 Kj ?
it will be 48.3 KJ.
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