How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 to 100.0

Q = 19255.6 j

Explanation:

Specific heat capacity:

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

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

Given data:

Mass of water = 46 g

change in temperature = ΔT = 100-0.0 = 100 °C

Heat absorbed by water = ?

Solution:

Specific heat capacity of water = 4.186 j/g. °C

Q = m.c. ΔT

Q = 46 g×4.186 j/g. °C×100 °C

Q = 19255.6 j

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