Work needed to heat 1 L of water by 10°C with falling mass

[Dars]

Homework Statement

You want to heat 1 L of water by 10 ºC, and so you heat it with a mechanical egg beater that is powered by a 1 kg mass on a rope over a pulley. How far does the mass have to descend to supply enough work for heating the water?

Homework Equations

|work| = mg(x_f - x_0)
C_P (liquid water) = 75.3JK^-1mol^1

The Attempt at a Solution

I first asked myself how much energy does it take to increase the temp of 1kg of water by 10degC.
Heat capacity of water (above) is the energy required to heat 1 mol of water by 1deg.
So I first determined how many moles of water i have:

1L = 1kg = 1000g * (1mol/18.01g) = 55.5mol of liquid water

Therefore 55.5mol *75.3JK^-1mol^1 = 4179.15J/K * 10K = 417915.0J

So it requires 417915.0J to raise the temp of 1kg water by 10degC.

This is the work required also because work is also in joules, --energy.

Then i used the formula for mechanical work to determine the value for delta x.

|work| = mg(x_f - x_0)

417915.0J(1kJ/10000J) = 1kg(9.81m/s/s)(x_f) (i got ride of x_0 because I set scale so that x_0 = 0)

x_f = 42.6m

So the mass has to descend through 42.6m in order to heat 1kg of liquid water by 10 degC.

Is this correct or if not please help..thanks a lot. I also have some other questions about the problem but i don't want to ask until i know if my answer is correct...thank you again

 

[Bystander]

1L = 1kg = 1000g * (1mol/18.01g) = 55.5mol of liquid water

Correct.

Therefore 55.5mol *75.3JK^-1mol^1 = 4179.15J/K * 10K = 417915.0J

"WORNG." Ten x four thousand equals four hundred thousand?

Then i used the formula for mechanical work to determine the value for delta x.
|work| = mg(x_f - x_0)
417915.0J(1kJ/10000J) = 1kg(9.81m/s/s)(x_f) (i got ride of x_0 because I set scale so that x_0 = 0)
x_f = 42.6m
So the mass has to descend through 42.6m in order to heat 1kg of liquid water by 10 degC.

"Four kilometers." You need to pay attention to decimal "orders of magnitude."