Question

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

Answer 1

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 :

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