A peice of ice weighting 5g at -20 centigrade is put into 10g of ice at 30C. Assuming no heat is lost to the outside.Calculate the final tempature of the mixture?
Ice at 30 deg C??? Not likely but anyway
specific heat of ice (about 2 J/gram deg c) does not matter in the end here
call it sh
heat into cold ice = heat out of hot ice
5 (T - -20) * sh = 10 (30-T) * sh
5 T + 100 = 300 - 10 T
15 T = 300
T = 20
BUT I doubt the whole thing
suspect that 10 grams was water at 30 deg
In that case
Heat in to heat the ice up to zero
5 (20)*sh
Heat in to melt the ice
5 * hm where hm is the heat of fusion of water
Heat in to warm melted ice up to T
5(T-0) * shw
where shw is specific heat of water
Heat out of 30 deg water
10(30-T)*shw
so
5 (20)*sh + 5 * hm + 5(T-0) * shw = 10(30-T)*shw
If T comes out below zero, then T = 0 and some of the 30 deg water froze
To calculate the final temperature of the mixture, we can use the concept of heat transfer and the principle of energy conservation.
The heat lost by the warm ice at 30°C will be equal to the heat gained by the cold ice at -20°C when they reach thermal equilibrium. The formula for heat transfer is:
Q = mcΔT
Where:
Q represents the heat transfer
m represents the mass of the substance
c represents the specific heat capacity of the substance
ΔT represents the change in temperature
Let's apply this formula to the problem:
1. Calculate the heat lost by the warm ice (30°C):
Q1 = m1 * c * ΔT1
Given:
m1 = 10g (mass of warm ice)
c = specific heat capacity of ice ~ 2.09 J/g°C (average value)
ΔT1 = final temperature - initial temperature
ΔT1 = final temperature - 30°C
2. Calculate the heat gained by the cold ice (-20°C):
Q2 = m2 * c * ΔT2
Given:
m2 = 5g (mass of cold ice)
c = specific heat capacity of ice ~ 2.09 J/g°C (average value)
ΔT2 = final temperature - initial temperature
ΔT2 = final temperature - (-20°C)
Since no heat is lost to the outside, the heat lost by the warm ice (Q1) is equal to the heat gained by the cold ice (Q2):
Q1 = Q2
Therefore:
m1 * c * ΔT1 = m2 * c * ΔT2
Substituting the given values:
10g * 2.09 J/g°C * (final temperature - 30°C) = 5g * 2.09 J/g°C * (final temperature - (-20°C))
Simplifying:
20.9 J/°C * (final temperature - 30°C) = 10.45 J/°C * (final temperature + 20°C)
20.9 * final temperature - 20.9 * 30 = 10.45 * final temperature + 10.45 * 20
20.9 * final temperature - 627 = 10.45 * final temperature + 209
10.45 * final temperature - 20.9 * final temperature = 209 + 627
-10.45 * final temperature = 836
final temperature = -836 / 10.45
final temperature = -80°C
Therefore, the final temperature of the mixture is -80°C.