Question

A sample of water is cooled from 45°C to 25°C

Answer 1

Answer:

1000 g

Explanation:

Given data:

Initial temperature of water = 45°C

Final temperature of water = 25°C

Heat released = 20 cal (4184×20 = 83680 j)

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 is 4.184 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 = 25°C - 45°C

ΔT = -20°C

83680 j = m × 4.184 j/g.°C ×20°C

83680 j = m × 83.68 j/g

m = 83680 j / 83.68 j/g

m = 1000 g