- #1
shin777
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30g of ice at -15 degree celcius is added to 100g of water at 25 degree celcius. What is the final temperature?
T1=ice
T2=H2O
at equilibrium must have
m1c1(Te-T1)+m1k=m2c2(T2-Te) [k=3.3x10^5 J/K fusion heat coeff.]
then
Te=(m1c1T1+m2c2T2-m1k)/(m1c1+m2c2)
T1=268 k
T2=298 K
c1=2090 J/(kg K)
c2=4186 J/(kg K)
Te=(0.02*2090*268+0.1*4186*298-0.02*3.3… / (0.02*2090+0.1*4186)
Te=280,9 K
Te=7.8 C
Final Temperature is 7.8 C
does this look ok?
T1=ice
T2=H2O
at equilibrium must have
m1c1(Te-T1)+m1k=m2c2(T2-Te) [k=3.3x10^5 J/K fusion heat coeff.]
then
Te=(m1c1T1+m2c2T2-m1k)/(m1c1+m2c2)
T1=268 k
T2=298 K
c1=2090 J/(kg K)
c2=4186 J/(kg K)
Te=(0.02*2090*268+0.1*4186*298-0.02*3.3… / (0.02*2090+0.1*4186)
Te=280,9 K
Te=7.8 C
Final Temperature is 7.8 C
does this look ok?