Find Lv for Calorimeter Problem

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To find the latent heat of vaporization (Lv) for steam in the calorimeter problem, the equation mLv + mc(ΔT) = mc(ΔT) + mc(ΔT) is used, where m represents mass and c represents specific heat. Given the masses and temperature changes of the calorimeter, water, and steam, the calculation involves substituting these values into the equation. The attempt at a solution resulted in a value of 4964.3, but the correctness of this result is uncertain. Clarification on the definition of Lv is also requested, indicating a need for further explanation. The discussion highlights the importance of understanding the variables and their relationships in solving calorimetry problems.
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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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