Question

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.​

Answer 1

mass of water = 1540 g

Step-by-step 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