Find Lv for Calorimeter Problem

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Homework Statement

calorimeter mass = 60 g
calorimeter and water mass = 290 g
temperature of cal. and water = 30 C
temp of steam = 90 C
mass of cal, water, and steam = 300 g
specific heat of cal = .1 cal/g C
final temp of cal, water, and steam = 80 C
Find the Lv for this steam.

Homework Equations

(a space between the variables means multiply)
m Lv + m c (change in temp) = m c (change in temp) + m c (change in temp)
m Lv is steam, the 1st m c T is steam water, the 2nd is water, and the 3rd is the cal, and i actually have no idea what Lv is...

The Attempt at a Solution

10 x Lv + 10 2.01 -10 = 230 4.19 50 + 60 (.1 x 4.19) 50
i came up with 4964.3, but i don't know if it is right, or if i even did it right.

 

[Mentz114]

You have not described the scenario, just a list of values. I don't know what Lv is either, that's for you to tell us.

 

[m2003]

I would approach this problem by first identifying the relevant variables and using the appropriate equations to solve for Lv, which represents the latent heat of vaporization for the steam in this calorimeter problem.

The relevant variables in this problem are the mass (m) and specific heat (c) of the calorimeter, water, and steam, as well as the initial and final temperatures of the system.

Using the equation Q = m c ΔT, where Q represents the heat transferred, m represents the mass, c represents the specific heat, and ΔT represents the change in temperature, we can set up the following equation:

Qsteam + Qwater + Qcal = 0

Since the calorimeter is insulated and no heat is lost to the surroundings, the total heat transfer must be equal to zero.

We can then substitute in the values given in the problem to solve for Lv:

msteam csteam (ΔTsteam) + mwater cwater (ΔTwater) + mcal ccal (ΔTcal) = 0

(300 g)(4.19 J/g°C)(90°C - 80°C) + (290 g)(4.19 J/g°C)(90°C - 30°C) + (60 g)(0.1 cal/g°C)(90°C - 80°C) = 0

Simplifying and converting the units to J/g, we get:

1257 J + 10923 J + 50 J = 0

Solving for Lv, we get:

Lv = -(1257 J + 10923 J + 50 J) / (300 g) = - 38.4 J/g

Therefore, the latent heat of vaporization for the steam in this calorimeter problem is 38.4 J/g. This value is negative because heat is being transferred from the steam to the water and calorimeter, causing the steam to condense and release its latent heat.