Energy lost in conversion of steam to ice

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Homework Help Overview

The discussion revolves around calculating the energy required to convert steam at a high temperature into ice at 0°C. The problem involves specific heat capacities and phase changes, focusing on the energy exchanges during cooling and freezing processes.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various calculations involving specific heat, latent heat of vaporization, and fusion. There are attempts to account for energy changes during cooling and phase transitions, with some participants questioning the accuracy of their calculations and the need to include all relevant energy exchanges.

Discussion Status

Multiple approaches to the problem are being discussed, with participants providing different calculations and questioning the inclusion of certain energy components. Some guidance has been offered regarding the specific heat of steam, but no consensus has been reached on the final energy value.

Contextual Notes

Participants are working with specific values for heat capacities and latent heats, and there is a mention of a potential typo regarding the heat capacity of steam. The original poster's calculations seem to be incomplete, as they do not account for all necessary energy changes.

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Homework Statement



How many J of energy must be removed when 145.0 g of steam, at a temperature of 188.0°C, is cooled and frozen into 145.0 g of ice at 0°C? Take the specific heat of steam to be 2.1 kJ/(kg·K).


Homework Equations



Q=m*L

The Attempt at a Solution



Energy = Mgas:liquid*Lvaporization + Mliquid:solid * Lfusion

.145*22.6E5 + .145*33.5E4 = 3.76E5 J
 
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You also need to account for the energy to cool the steam from 188C to and then cool the water from 100C to 0C.
Use Q = m*c*T
 
hm... it's still not working.

I added .145(2.1)(88) + .145*4186*100 = 60723.769 J

The total I found was 4.37E5 J.
 
cool steam = 0.145 * (188-100) * 2100
condense steam = 0.145 * 2272E3
cool water = 0.145 * (100-0) * 4181
freeze water = 0.145 * 334E3

Just a small typo, the heat capcity of steam is 2.1KJ/kg = 2100 J/kg
 

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