Calculating Unfrozen Water After Heat Transfer

[Pinkythermo]

Homework Statement

How much water remains unfrozen after 50.2kJ is transferred as heat from 260g of liquid water initially at its freezing point?

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

Homework Equations

What equations do you think you will need here? Sensible heating maybe?

The Attempt at a Solution

You need to show some attempt before anyone at PF can help you.

 

[Pinkythermo]

k may b these equations

Q =mc(Tf - Ti)
Q =+/-mL

 

[rock.freak667]

Q =+/-mL

So how much mass is converted from ice to water?

 

[DeeNos]

To calculate the amount of water that remains unfrozen after heat transfer, we can use the specific heat capacity and latent heat of fusion of water. The specific heat capacity of water is 4.186 J/g°C and the latent heat of fusion is 334 J/g.

First, we need to calculate the amount of energy required to raise the temperature of 260g of water from its freezing point (0°C) to its melting point (0°C). This can be done using the equation Q = mCΔT, where Q is the energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Q = (260g)(4.186 J/g°C)(0°C - 0°C) = 0J

Next, we need to calculate the amount of energy required to melt the 260g of water. This can be done using the equation Q = mL, where Q is the energy, m is the mass, and L is the latent heat of fusion.

Q = (260g)(334 J/g) = 86,840J

Now, we can subtract the energy required for temperature change from the total energy transferred to get the amount of energy available for melting.

86,840J - 0J = 86,840J

Finally, we can convert the energy into kilojoules and use it to calculate the amount of water melted.

86,840J = 86.84kJ

The amount of water melted can be calculated using the equation m = Q/L, where m is the mass, Q is the energy, and L is the latent heat of fusion.

m = (86.84kJ)/(334 J/g) = 0.26g

Therefore, after 50.2kJ of heat is transferred, 0.26g of water will remain unfrozen. This can also be expressed as a percentage, where (0.26g/260g) x 100% = 0.1% of the initial amount of water remains unfrozen.