What is the final temperature of the aluminum can after heating water?

AI Thread Summary
The discussion revolves around calculating the final temperature of an aluminum can after heating water. A 1.0 g sample of water at 28 degrees Celsius is placed in a 200 g aluminum can initially at 300 degrees Celsius. The specific heat capacities of aluminum and water, along with the latent heat of vaporization for water, are provided for the calculations. The approach involves determining the heat required to raise the water's temperature to 100 degrees Celsius and then the heat needed to vaporize it. The total heat absorbed by the water equals the heat lost by the aluminum can, leading to the final temperature calculation.
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Homework Statement



1.0 g of water at 28 degree celcius is placed into a 200g aluminium can at 300 degree celcius. What is the temperature of the aluminium can after all the water vapourises? Given the specific heat capacity of aluminium, 900 J/kg/degree celcius and water, 4200 and specific latent heat of water, 2.26 x 10^6 J/kg assuming no heat is loss to the surrounding.

Homework Equations





The Attempt at a Solution



I am not sure how to go about when it comes to combined specific heat capacity. Do we simply add them up?
 
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Find the amount of heat needed to rise the temperature of 1 g of water from 28 degree celsius to 100 degree celsius.
Next find amount of heat needed to evaporate 1 g of water from 100 degree celsius to 1 g of steam at 100 degree celsius. Find the total amount of heat.
This much heat is lost by the aluminum can.
 
rl.bhat said:
Find the amount of heat needed to rise the temperature of 1 g of water from 28 degree celsius to 100 degree celsius.
Next find amount of heat needed to evaporate 1 g of water from 100 degree celsius to 1 g of steam at 100 degree celsius. Find the total amount of heat.
This much heat is lost by the aluminum can.

thank you.
 
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