Adding Ice to Water?

Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.240 kg and a temperature of 65.9 C.
How much ice at a temperature of -15.7 C must be dropped into the water so that the final temperature of the system will be 33.0 C?
Take the specific heat of liquid water to be 4190 J/kg K, the specific heat of ice to be 2100 J/kg K, and the heat of fusion for water to be 334 kJ/kg.

Answer 1

You want the final temp to be 33C, which is above freezing.

The energy in the system will be equal, nothing is lost.

i.e. the energy required to raise the ice to 0C, plus to melt the ice, plus to raise the water from the ice to 33C (the energy required to raise the temp or melt the ice) must equal the energy released by the cooling of the water in the container.

The formula for the energy required to change the temperature of a substance is:

q=m*c*(change in temp.), where q=energy, m=mass, c=specific heat, and in the parenthesis is the change in temperature the substance goes through.

The formula for the energy required to change ice from solid to liquid is:

q=m*Hf, where Hf=heat of fusion.

Step 1, calculate the energy released when the water cools:

so, the water goes from 65.9C to 33C, a temp change of 65.9-33=32.9C
q=m*c*(change in temp) becomes
q=0.240*4190*32.9, which is q=33084.24 released

Step 2, figure the equations for the energy required to raise the temp of the ice and to change the ice to water, and to raise the temp of the water from the ice and add them together (this will be equal to the energy released in step 1)

The ice goes from -15.7C to 32C, a temp change of 47.7C
q=m*c*(change in temp) becomes
q=m*2100*47.7, or q=m*100170

The ice then goes from solid to liquid
q=m*Hf, which becomes q=m*334

The water from the ice then goes from 32C to 33C, a temp change of 1C
q=m*c*(change in temp) becomes
q=m*4190*1, which becomes q=m*4190

All this q added together is the energy reguired to raise the ice to 33C, which must be equal to the energy released from the first step, so:

m*100170+m*334+m*4190=33084.24 , so:

m(100170+334+4190)=33084.24 , so:

m(104694)=33084.24 , so:

m=33084.24/104694 ,so:

m=.316 kg

Answer 2

Heat balance problem. The ice must first heat up from -15.7C to 0C, then it must melt. The heat consumed by this will cool the liquid water. You can calculate how much heat needs to be removed first with the mass of water, the delta T (temperature change), and the specific heat of liquid water.
Then figure out how much ice is required to consume that much heat in the two steps: heat up to 0C (specific heat of ice) and then melting (latent ht of fusion) Good luck

Answer 3

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Answer 4

kev, somewhere in your calculations you are wrong. Double check it.