-25.116 Kilojoules are given off when 120 grams of water are cooled from 25°C to -25°C.
Explanation:
GIVEN THAT:
mass of water = 120 grams OR 0.12 Kg
initial temperature = 25 degrees
final temperature = -25degrees
change in temperature (ΔT) = final temperature - initial temperature
-25 -25
= -50 degrees
specific heat capacity of water = 4.186 joule/gram degree Celsius
Q (heat energy transferred) = ?
Formula used,
Q = mcΔT
Putting the values in equation
Q = 120 X 4.186 X -50
= -25116 joules
The energy in joules is converted to kilo joule by dividing it with 1000.
The negative sign in the energy transfer shows that energy is released in the process.
