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hils0005
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A 10g ice cube at -10C is placed in a lake whose temp is 15C, calculate the change in entropy of the cube-lake system as the ice comes to thermal equilibrium with the lake.
specific heat of ice is 2220 J/KgK
Q(released)=Q(absorbed)
U=Q-W
S=Q/T
L(f)water=333kJ/kg
Not really sure how to start??
Do you need to calculate change in the ice from -10C to 0c, then from 0C ice to 0C water, 0C water to 15C?
1. Q(ice)=c(ice)M(ice)(Tf-Ti)
Q=(2220 J/kgK)(0.01kg)(273K-263K)=222J
2.Q(ice to water)=Lm=333kJ/kg(.01kg)=3.33kJ=3330J ?
3.Q(water)=c(w)m(w)(Tf-Ti)
Q=(4190J/kgK)(.01kg)(288K-273K)=628.5J
Q=222+3330+628.5=4180.5J
I don't know what to do now or even know if I'm on the right path
specific heat of ice is 2220 J/KgK
Homework Equations
Q(released)=Q(absorbed)
U=Q-W
S=Q/T
L(f)water=333kJ/kg
The Attempt at a Solution
Not really sure how to start??
Do you need to calculate change in the ice from -10C to 0c, then from 0C ice to 0C water, 0C water to 15C?
1. Q(ice)=c(ice)M(ice)(Tf-Ti)
Q=(2220 J/kgK)(0.01kg)(273K-263K)=222J
2.Q(ice to water)=Lm=333kJ/kg(.01kg)=3.33kJ=3330J ?
3.Q(water)=c(w)m(w)(Tf-Ti)
Q=(4190J/kgK)(.01kg)(288K-273K)=628.5J
Q=222+3330+628.5=4180.5J
I don't know what to do now or even know if I'm on the right path