Solve Thermodynamics: 7.67 kg Lead & 764 g Water Boiling Pt

[al86]

Homework Statement

Suppose 7.67 kg of molten (liquid) lead at its melting point is poured into 764 g of water at 53.3o C. How much of the water boils?

Homework Equations

mc(detla T) = m(water)c(detla T).

The Attempt at a Solution

i m guessing i have to do the dollowing:
mc(detla T) = m(water)c(detla T)+m(vapor)Lv
am i on the right track..
thanks in advance

 

[turin]

describe the physical change that occurs in the lead.

 

[thomate1]

.

I would first clarify that the problem is asking for how much water boils, rather than how much lead boils. Based on the given information, it seems that the lead is being poured into the water, which would cause the water to boil and not the lead.

To solve this problem, we can use the formula mc(delta T) = m(water)c(delta T) + m(vapor)Lv, where m is the mass, c is the specific heat capacity, delta T is the change in temperature, and Lv is the latent heat of vaporization.

First, we can calculate the change in temperature for the water using the equation Q = mc(delta T), where Q is the heat transferred. We know that the lead is at its melting point, so it will transfer heat to the water until both substances reach the same temperature. Therefore, we can set up the equation as follows:

m(lead)c(lead)(Tf - Tm) = m(water)c(water)(Tf - 53.3)

Where m(lead) = 7.67 kg, c(lead) is the specific heat capacity of lead, Tf is the final temperature (unknown), Tm is the melting point of lead, m(water) = 764 g, c(water) is the specific heat capacity of water, and 53.3 is the initial temperature of the water.

Next, we can calculate the mass of water that will boil using the formula m(vapor) = Q/Lv, where Q is the heat transferred and Lv is the latent heat of vaporization for water. We can use the same value for Q that we calculated in the previous step.

Finally, we can use the mass of water that boils to calculate the volume of water that boils using the density of water.

In conclusion, using the given information and thermodynamic principles, we can calculate the amount of water that will boil when 7.67 kg of molten lead is poured into 764 g of water at 53.3o C.