Question 224.
Suppose that in an insulated container, 0.100 kg of water at 20.0 degrees celcius is mixed with 1.500 kg of ice at -15.0 degrees celcius. Find the final temperature of the system.
Answer 224.
1)
100 g of water will give up 2000 cal of heat in cooling from 20° C to 0 °C.
2)
1500 g of ice will need 1500 * 0.5 * 15 = 11250 cal of heat to get heated up from
-15 °C to 0 °C
3)
Since heat needed by subcooled ice is much more than what water can give by cooling down to 0 °C, some water will have to freeze to ice to provide that heat.
4)
2000 cal of heat is given by water cooling to 0 °C and further 9250 cal of heat is needed for entire quantity of ice to get heated to 0 °C. 100 g os water can give 80 * 100 = 8000 cal of heat if it waere to freeze to ice at 0 °C.
5)
From the above observations, it is clear that the quantity of ice which is subcooled to - 15 °C will not heat up to 0 °C as the heat available from 100 g water will be insufficient. Suppose, ice gets heated up to - T °C, then
heat needed by 1500 g of ice = 1500 * 0.5 * (15 - T)= 750 (15 - T)
6)
With heat balance in the above situation,
750 (15 - T) = 2000 + 8000 + 100 * T
=> 11250 - 750T = 10000 + 100 * 0.5 *T
=> 800T = 1250
=> T = 1.5625 °C
That means the final temperature of the mixutre will be - 1.5625 °C and the entire mixture will be subcooled ice at - 1.5625 °C ======================================
Verification:
Heat lost by 1500 g of ice = 1500 * 0.5 * (15 - 1.5625) = 10078 cal
Heat gained by water at 20 °C to get frozen to subcooled ice at - 1.5625 °C
= 100 * (20 + 80 + 0.5 * 1.5625) cal = 10078 cal.
Suppose that in an insulated container, 0.100 kg of water at 20.0 degrees celcius is mixed with 1.500 kg of ice at -15.0 degrees celcius. Find the final temperature of the system.
Answer 224.
1)
100 g of water will give up 2000 cal of heat in cooling from 20° C to 0 °C.
2)
1500 g of ice will need 1500 * 0.5 * 15 = 11250 cal of heat to get heated up from
-15 °C to 0 °C
3)
Since heat needed by subcooled ice is much more than what water can give by cooling down to 0 °C, some water will have to freeze to ice to provide that heat.
4)
2000 cal of heat is given by water cooling to 0 °C and further 9250 cal of heat is needed for entire quantity of ice to get heated to 0 °C. 100 g os water can give 80 * 100 = 8000 cal of heat if it waere to freeze to ice at 0 °C.
5)
From the above observations, it is clear that the quantity of ice which is subcooled to - 15 °C will not heat up to 0 °C as the heat available from 100 g water will be insufficient. Suppose, ice gets heated up to - T °C, then
heat needed by 1500 g of ice = 1500 * 0.5 * (15 - T)= 750 (15 - T)
6)
With heat balance in the above situation,
750 (15 - T) = 2000 + 8000 + 100 * T
=> 11250 - 750T = 10000 + 100 * 0.5 *T
=> 800T = 1250
=> T = 1.5625 °C
That means the final temperature of the mixutre will be - 1.5625 °C and the entire mixture will be subcooled ice at - 1.5625 °C ======================================
Verification:
Heat lost by 1500 g of ice = 1500 * 0.5 * (15 - 1.5625) = 10078 cal
Heat gained by water at 20 °C to get frozen to subcooled ice at - 1.5625 °C
= 100 * (20 + 80 + 0.5 * 1.5625) cal = 10078 cal.
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