Question

A sample of water is cooled from 45°C to 35°C by the removal of 84 Joules of heat. What is

the mass of the water?

Answer 1

m = 2 g

Step-by-step explanation:

Given data:

Initial temperature = 45°C

Final temperature = 35°C

Heat evolved = 84 j

Mass of water = ?

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

specific heat capacity of water is 4.18 J/g.°C

ΔT = 35°C - 45°C

ΔT = - 10°C

84 j = m ×4.18 J/g.°C ×- 10°C

84 j = m × -41.8 J/g

m = 84 j / -41.8 J/g

m = 2 g

negative sign shows heat is evolved.