Calculate the amount of energy absorbed by 45.0 g sample of water to raise its temperature from 18.0C to 48.0 C. The specific heat of water is 4.18 J/g C. 1000 J= 1kj

5.643 kJ

Explanation:

The quantity of heat released or absorbed by a substance (Q) is given by the equation:

Q = mcΔT

Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.

Given that:

m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C

ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C

The quantity of heat is:

Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J

Q = 5.643 kJ

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