Adding Ice to Water An insulated beaker with negligible mass contains liquid wat

Question

Adding Ice to Water

An insulated beaker with negligible mass contains liquid water with a mass of 0.295 kg and a temperature of 73.3 C .

Part A

How much ice at a temperature of -14.9 C must be dropped into the water so that the final temperature of the system will be 24.0C ?

Take the specific heat of liquid water to be 4190 J/kgK , the specific heat of ice to be 2100 J/kgK , and the heat of fusion for water to be 3.34×105 J/kg .

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Adding Ice to Water

An insulated beaker with negligible mass contains liquid water with a mass of 0.295 kg and a temperature of 73.3 C .

Part A

How much ice at a temperature of -14.9 C must be dropped into the water so that the final temperature of the system will be 24.0C ?

Take the specific heat of liquid water to be 4190 J/kgK , the specific heat of ice to be 2100 J/kgK , and the heat of fusion for water to be 3.34×105 J/kg .

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Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

The amount of heat that will be lost (which will be equal to the amount of heat absorbed by ice) is
= mass*specific heat of water*delta(temperature)
=0.295*4190*49.3=60937.2 Joules
The amount of heat absorbed by ice
= mass*specific heat of ice*delta(temperature)
60937.2 = m*2100*38.9
m =60937.2/ 81690= 0.746 kg
I didn't convert Celsius into kelvin because the differences are the same .

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