timot0617
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Two part question, I can answer the first one, but i don't know how
A 200g calorimeter made of aluminum contains 500g of water at 20°C. A 100g piece of ice is cooled to -20°C and is placed in the calorimeter.
(a)Find the final temperature of the system, assuming no heat loss.
(b)If we are to add a second 200g piece of ice at -20°C, how much ice remains in the system after it reaches equilibrium?
Some Constants and Equations
Specific Heat of Ice=2.0kJ/kg°C
Specific Heat of Water=4.186kJ/kg°C
Specific Heat of Aluminum=.91kJ/kg°C
Q=mcΔT and Q=mL
I answered the first one(a) and got 16.77°C.
I tried to answer the second one(b)
Q(1)=mcΔT=(.2kg)(2.0kJ/kg°C)(0°C-(-20°C)) [From -20°C to 0°C]
Q(2)=mL=(.2kg)(334kJ/kg)=66.8kJ [Ice-To-Water]
I'm stuck here... Also, am I doing it right?
Thanks for any help.
A 200g calorimeter made of aluminum contains 500g of water at 20°C. A 100g piece of ice is cooled to -20°C and is placed in the calorimeter.
(a)Find the final temperature of the system, assuming no heat loss.
(b)If we are to add a second 200g piece of ice at -20°C, how much ice remains in the system after it reaches equilibrium?
Some Constants and Equations
Specific Heat of Ice=2.0kJ/kg°C
Specific Heat of Water=4.186kJ/kg°C
Specific Heat of Aluminum=.91kJ/kg°C
Q=mcΔT and Q=mL
I answered the first one(a) and got 16.77°C.
I tried to answer the second one(b)
Q(1)=mcΔT=(.2kg)(2.0kJ/kg°C)(0°C-(-20°C)) [From -20°C to 0°C]
Q(2)=mL=(.2kg)(334kJ/kg)=66.8kJ [Ice-To-Water]
I'm stuck here... Also, am I doing it right?
Thanks for any help.