A single 50g ice cube is dropped into a thermally insulated container holding 200g of water. The water is initially at 25oC and the ice is initially at -15oC.(adsbygoogle = window.adsbygoogle || []).push({});

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a) What is the final temperature of the system after is has come to thermal equilibrium ?

b) Now lets drop a second 50gm cube of ice into the system. What is the final temperature of the system after it has come to thermal equilibrium for this second time?

c) What is the total mass of ice melted during this entire process?

part a)

Basically I figured out total heat absorbed by ice, which is equal to total heat lost by the water. With some algebra, I solved for Temperature, and the new temperature is now 2.5155 C.

Q1=m*c*delta T

= .05kg*specific heat ice*15 C

=1665J

Q2= L*m

= 333,000J/kg*.05kg =16650 J

Q3= mass*specific heat water*delta T

0.05kg*4190*T

= 209.5T

Q1+Q2+Q3=Q_water

where Q_water = .2*4190*(25-T)

T=2.5155 degrees C

part b) Okay, the final temp is now the initial temp for my water system..How do I change my equations to reflect the addition of the 2nd ice cube?

part c)

How do I go about finding total mass of ice melted?

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# Simple thermo dynamics help

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