7. A certain mass of water was heated with 41,840 Joules, raising its
temperature from 22.0 °C to 28.5 °C. Find the mass of water.

mass of water = 1540 g

Explanation:

Given data:

Heat absorbed = 41840 j

Initial temperature = 22.0°C

Final temperature = 28.5 °C

Mass of water = ?

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 = 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 = 28.5 °C - 22 °C

ΔT =  6.5°C

Now we will  put the values in formula:

41840 J = m ×4.18 j/g.°C× 6.5°C

41840 J = m × 27.17 j/g

m = 41840 J /27.17 j/g

m = 1540 g

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explanation:

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