Find Lv for Calorimeter Problem

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SUMMARY

The discussion focuses on calculating the latent heat of vaporization (Lv) for steam in a calorimetry problem involving a calorimeter, water, and steam. The provided data includes a calorimeter mass of 60 g, a combined mass of 290 g for the calorimeter and water, and a final temperature of 80 °C. The equation used is mLv + mc(ΔT) = mc(ΔT) + mc(ΔT), where the user attempts to solve for Lv but expresses uncertainty about the calculations and the meaning of Lv. The final value calculated by the user is 4964.3, though its accuracy is questioned.

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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.
 
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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.
 

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