How many joules are needed to change the temperate of 22g of water from 18°C to 33°C?

Q =  1379.4 J

Explanation:

Given data:

Mass of water = 22  g

Initial temperature = 18°C

Final temperature = 33°C

Heat absorbed = ?

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 = 33°C - 18 °C

ΔT =  15°C

Q = 522 g ×4.18 J/g.°C× 15°C

Q =  1379.4 J

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