When 0.15 kg of ice at 0 degree celsius is mixed with 0.30 kg of water at 50 degree celsius in a container. Then find the temperature of the mixture?
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In this question we use the concept of specific heat capacity to obtain the latent heat of fusion of ice.
Heat gained by ice = heat lost by the water.
The change in temperature for the water is:
50 - 6.7 =43.3°C
The change in temperature of ice is:
6.7°C
Q=mcΔT where Q is the quantity of heat gained or lost, m the mass of the substance, c the specific heat capacity of water and ΔT the temperature change.
Heat gained by ice is: mLf + mcΔT where lf is the latent heat of fusion and c the specific heat capacity of water.
In this question we are looking for the Lf thus it is not known.
Substituting what we is given in the formulae we get :
0.15Lf + 0.15 × 4186 × 6.7 =0.15Lf + 4206.93
Heat lost by the water is:
0.30 ×4186 × 43.3 =54376.14Joules
54376.14 = 0.15Lf + 4206.93
0.15Lf = 54376.14 - 4206.93
0.15Lf = 50169.21
"LF"=50169.21 ÷ 0.15 =334461.40joules/Kg
____________________
_____________________
In this question we use the concept of specific heat capacity to obtain the latent heat of fusion of ice.
Heat gained by ice = heat lost by the water.
The change in temperature for the water is:
50 - 6.7 =43.3°C
The change in temperature of ice is:
6.7°C
Q=mcΔT where Q is the quantity of heat gained or lost, m the mass of the substance, c the specific heat capacity of water and ΔT the temperature change.
Heat gained by ice is: mLf + mcΔT where lf is the latent heat of fusion and c the specific heat capacity of water.
In this question we are looking for the Lf thus it is not known.
Substituting what we is given in the formulae we get :
0.15Lf + 0.15 × 4186 × 6.7 =0.15Lf + 4206.93
Heat lost by the water is:
0.30 ×4186 × 43.3 =54376.14Joules
54376.14 = 0.15Lf + 4206.93
0.15Lf = 54376.14 - 4206.93
0.15Lf = 50169.21
"LF"=50169.21 ÷ 0.15 =334461.40joules/Kg
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