Question

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

Answer 1

75240J

Step-by-step 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.