Latent Heat + Specific Heat Problem

[yaylee]

Homework Statement

Suppose molten (liquid) lead, mass ml = 9.83 kg, is at its melting point. The lead is poured into water of mass mw = 585 g and initial temperature Tw = 16.2 C. Find mwb, the amount of the water that boils.
NOTE: Reaching the boiling point is not enough; the question asks for the amount of water that vaporizes as well.
ASSUME: no water evaporates; it only vaporizes at the boiling point.

Homework Equations

Constants: M.P Lead = 327 C, MP Water = 100 C
Latent Heat Lead = 33 kJ/kg
Latent Heat Q = mL
Specific Heat Eq'n: Q=mcΔT

The Attempt at a Solution

mL + mcΔT = mcΔT + mL
9.83 kg (33 kJ/kg) + 9.83 kg(0.130 kJ/kg)(327C - 100C) = 0.585(4.186)(100 -16.2) + m(2260 kJ/kg)
Solving for m, we get m = 0.215 kg.

Did I go wrong in my reasoning here? Many thanks in advance for your help!

 

[TSny]

Your equation looks good to me. But I don't get the same numerical answer.

 

[NasuSama]

Your equation looks good to me. But I don't get the same numerical answer.

What do you get when you plug and chug in the values? I worked on the similar type of problem, and I got stuck on it from using that equation.

 

[TSny]

I get about 0.181 kg. What do you get for the numerical value of the left side of the equation? What do you get for the simplified form of the right side?

 

[Nickie]

Your reasoning and calculations seem to be correct. However, it is important to note that the latent heat of vaporization for water is different from the latent heat of fusion for lead. The correct equation to use for this problem would be:

mL + mw(2260 kJ/kg) = mwb(2260 kJ/kg) + mw(4.186 kJ/kg)(100 - 16.2)

Solving for mwb, we get mwb = 0.210 kg, which is slightly different from your answer of 0.215 kg. This could be due to rounding errors or using different values for the specific heat of water. Overall, your approach and reasoning were correct, but make sure to use the correct equations and values for the specific heat and latent heat of the substances involved.