Question

5. To make ice, water must first be cooled.

The specific heat of water is 4,186 J/kg . •C.
Approximately how much heat must be removed from 0.50kg of water to change its
temperature from 24-C to 5-C? (2 points)
02
O 19,900)
O 39,800 ]
79,500)

Answer 1

Q = -39767 J

Step-by-step explanation:

Given data:

Specific heat of water = 4186 J/Kg.°C

Mass of water = 0.50 Kg

Initial temperature = 25°C

Final temperature = 5°C

Heat released = ?

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

ΔT = 5°C - 24°C

ΔT = -19°C

Q = 0.50 Kg×4186 J/Kg.°C× -19°C

Q = -39767 J