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How much heat is released as the temperature of 465 g of water drops from 75.0°C to 18.2°C?

User Omnigazer
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1 Answer

4 votes

The heat released : -110560.632 J

Further explanation

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Known

m =465 g

c water = 4.186 J/gram °C

∆T = temperature difference = 18.2 - 75 = - 56.8 °C

Then the heat :


\tt Q=m.c.\Delta T\\\\Q=465* 4.186* -56.8\\\\Q=\boxed{\bold{-110560.632~J}}\rightarrow -=exothermic

User Rubayeet
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