Thermodynamics: Calorimeter problem

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SUMMARY

The calorimeter problem involves a 50.0g copper calorimeter containing 250g of water at 20°C, requiring the calculation of steam mass needed to reach a final temperature of 50°C. The equation used is Qsteam = Qcalorimeter + Qwater, where ms * Ls represents the heat lost by steam. The correct approach involves considering both the latent heat of condensation (2.26 x 10^6 J/kg) and the heat lost as the steam cools from 100°C to 50°C. The calculated heat flow results in a total of 7638.6 cal needed for the system.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically heat transfer.
  • Familiarity with the concept of latent heat, particularly latent heat of condensation.
  • Knowledge of specific heat capacities, including those of water and copper.
  • Ability to manipulate equations involving mass, temperature change, and heat transfer.
NEXT STEPS
  • Study the calculation of latent heat in thermodynamics.
  • Learn how to apply the specific heat formula in heat transfer problems.
  • Explore the concept of heat flow in phase changes, particularly in steam and water.
  • Investigate the principles of calorimetry and its applications in thermal energy calculations.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling calorimetry problems, as well as educators and professionals in physics and engineering fields who require a solid understanding of heat transfer and phase changes.

petefic
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Homework Statement


A 50.0g copper calorimeter contains 250g of water at 20 deg celsius. How much steam must be condensed into the water if the final temperature of the system is to reach 50 deg celsius?

Homework Equations


Can someone explain how to find the Ls? I was told that just plugging in the latent heat of steam (2.26*10^6) is incorrect.

The Attempt at a Solution



Qsteam = Qcalorimeter + Qwater
ms * Ls = (mc * cc * dT) + (mw * cw * dT)
ms * Ls = (0.05kg * 0.0924cal/g * (50deg-20deg)) + (.25kg * 1cal/g *(50deg-20deg))
ms * Ls = 7638.6cal
 
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petefic said:

Homework Statement


A 50.0g copper calorimeter contains 250g of water at 20 deg celsius. How much steam must be condensed into the water if the final temperature of the system is to reach 50 deg celsius?

Homework Equations


Can someone explain how to find the Ls? I was told that just plugging in the latent heat of steam (2.26*10^6) is incorrect.

The Attempt at a Solution



Qsteam = Qcalorimeter + Qwater
ms * Ls = (mc * cc * dT) + (mw * cw * dT)
ms * Ls = (0.05kg * 0.0924cal/g * (50deg-20deg)) + (.25kg * 1cal/g *(50deg-20deg))
ms * Ls = 7638.6cal
You have to assume that the steam is at 100°C.

You have the right idea in setting up an equation that equates the heat flow into the 250 g of water and 50 g calorimeter to the heat flow out of an unknown mass of steam. But the heat flow from the steam occurs in two stages: 1. the release of the latent heat of condensation and 2. the heat lost in reducing its temperature from 100C to 50C.

AM
 

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