Download Heat Lost Heat Gained problems The heat lost by one substance in

Heat Lost Heat Gained problems
The heat lost by one substance in a system is gained by another in the system
when there is a difference in temperature between the substances. There is
also a heat transfer between the system and its surroundings if they are at
different temperatures.
Example:
100g of iron with a specific heat of 0.444J/g°C at 25°C is added to 50g of water
at 80°C. Find the final temperature.
Qlost=Qgained
gCpΔT= gCpΔT
(50g)(4.18J/g°C )(80°C-x)=(100g)( 0.444J/g°C)(x-25°C)
For ΔT we have to do some substitution. We don’t know the final
temperature and so we can’t directly calculate the change in temperature.
Therefore we choose x to be the variable that represents the final temperature.
We still need to plug in change in temperature to the equation. We want the
change in temperature to be a positive value. We know that the heat will
transfer from the higher temperature substance to the lower temperature
substance. The higher temperature will lose heat and the lower temperature
substance will gain heat. So 80 °C is a larger temperature than the final temp
x. So a positive change in temperature can be made by setting up the change in
temperature to be 80°C-x. The lower temperature substance (iron) will gain
heat. Therefore the final temperature x is a larger value than 25°C. The change
in temperature can be written as x-25°C for the iron.
Simplify the multiplication, distribute, combine like terms and solve for x.
(209)(80-x)=44.4(x-25)
16720-209x=44.4x-1110
16830=253.4x
x=66.42°C
Example 2:
If 80g of gold at 50°C is added to 10g of water at 25°C, the final
temperature rests at 30°C. Find the specific heat of gold in J/g°C.
gCpΔT= gCpΔT
(80g)(Cp)(20°C)=(10g)(4.18J/g°C)(5°C)
Cp=0.13J/g°C