Thermal Properties - Specific heat.

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SUMMARY

The discussion focuses on calculating the mass of steam required to raise the temperature of a calorimeter system containing water and ice from 0°C to 30°C. The specific heat capacities of copper, water, and ice are provided as 380 J/kg·K, 4200 J/kg·K, and 2100 J/kg·K, respectively. The specific latent heats of fusion and vaporization are given as 3.25 x 105 J/kg and 2.25 x 106 J/kg. The key equations used in the calculations are Q=mcΔθ for heat transfer and Q=ml for phase changes.

PREREQUISITES
  • Understanding of specific heat capacity and its application in thermal calculations.
  • Familiarity with the concepts of latent heat and phase changes.
  • Proficiency in using the equations Q=mcΔθ and Q=ml for thermal energy calculations.
  • Basic knowledge of calorimetry and heat transfer principles.
NEXT STEPS
  • Calculate the total heat required to raise the temperature of the water and ice to 30°C.
  • Determine the mass of steam needed by equating the heat gained by the water and ice to the heat lost by the steam.
  • Explore the effects of different initial temperatures on the mass of steam required.
  • Investigate the impact of varying the specific heat capacities of materials on thermal calculations.
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone involved in calorimetry or thermal energy calculations.

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Homework Statement


A well lagged calorimeter of mass 120g contains 200g of water and 50g of ice, initial at 0°C. A jet of steam is blown through the water until the water temperature reaches 30°C.

Calculate the mass of steam that must condensed.

The specific heat capacity of Copper: 380Jkg-1K-1
The specific heat capacity of Water: 4200Jkg-1K-1
The specific heat capacity of Ice: 2100Jkg-1K-1

The specific latent heat of fusion of Ice: 3.25x105 Jkg-1
The specific latent heat of vaporization of Water: 2.25x106Jkg-1


Homework Equations


Q=mcΔθ
Q=ml

The Attempt at a Solution


My attempt was using Q=mcΔθ, for water (200x-3) x (4200) x (30) = 25200Jkg-1K-1

And similar for Ice

(50x10-3)x(2100)x(30)=3150Jkg-1K-1


However to me it doesn't look like I'm going anywhere.. any help will be appreciated.
 
Physics news on Phys.org
Determine all the components that need to be warmed up. Calculate the total heat that needs to be added in order to bring all of those components to the final temperature.

Where's the heat going to come from?
 

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