How much heat would be released if 300 g of water cooled from 85°C to 25°C

75240J

Explanation:

Given parameters:

Mass of water  = 300g

Initial temperature of the water  = 85°C

Final temperature  = 25°C

Unknown:

Amount of heat released = ?

Solution:

The amount of heat released in this cooling process can be calculated using the expression below;

     H  = m x c x (t₂  - t₁)

m is the mass

c is the specific heat capacity of water  = 4.18J/g°C

t is the temperature

  Insert the parameters and solve;

 H = 300 x 4.18 x (25  - 85) = -75240J

The negative symbol denotes heat loss.

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