How many Joules of heat will be given off by 55.0g of water as it cools from 87.3°C to 25.0°C?

Q = -14322.77 J

Explanation:

Given data:

Mass of water = 55.0 g

Initial temperature = 87.3°C

Final temperature = 25.0 °C

Heat given off = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25.0 °C - 87.3°C

ΔT = - 62.3 °C

Q = 55.0 g×4.18 J/g.°C × - 62.3 °C

Q = -14322.77 J