Calculate the amount of energy that is absorbed by 75 g.. water
from 22.5 °c to 37.8 °c. the specific heat of water is 1 cal/ g x °c

Q = 1147.5 cal

Explanation:

Given data:

Energy absorbed = ?

Mass of water = 75 g

Initial temperature = 22.5°C

Final temperature = 37.8°C

Specific heat of water = 1 cal/g.°C

Solution:

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  = T2- T1

ΔT  = 37.8°C - 22.5°C

ΔT  = 15.3 °C

Q = m.c. ΔT

Q = 75 g . 1 cal/g.°C . 15.3 °C

Q = 1147.5 cal

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