Heat energy transferred in converting ice into steam

AI Thread Summary
To calculate the heat energy needed to convert 0.10 kg of ice at -10°C into steam at 100°C, multiple stages must be considered: heating the ice to 0°C, melting it into water, heating the water to 100°C, and finally vaporizing the water into steam. The calculations reveal that the total energy required is 301.6 kJ, which includes the specific heat capacities and latent heats for each phase change. The initial calculation underestimated the energy needed for vaporization due to a misprint in the latent heat value. It is crucial to double-check calculations to avoid significant errors. The discussion emphasizes the importance of understanding each phase change in thermal energy calculations.
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1. Homework Statement

Calculate the heat energy required to convert 0.10 kg of ice at -10 degrees Celcius into steam at 100 degrees Celcius.




2. Homework Equations

E = m c dt (1)

where

E = heat energy (kJ)

m = unit mass (kg)

c = specific heat capacity (kJ/kg oC)

dt = temperature change (Kelvin or degrees Celsius)





3. The Attempt at a Solution

Given in the question:

m = 0.1 kg
dt = 100-(-10) = 110 degrees Celcius
c = 2100 J Kg^-1 K^-1 (specific heat capacity of ice)


So what I did was using the equation:

E = m c dt

E = (0.1)(2100)(110)

E = 23100 J

which would be 23.1 kJ



The problem is that the answer section of the book says 301.6 KJ which is way above my result.
What's the mistake I'm doing?

Help appreciated.

 
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Ice must be heated to 0C; ice must then be converted into liquid water at 0C;
Water at0C must be heated to 100C; water must be converted to vapor (steam) at 100C.
There are 4 stages, each requiring heat energy
 
Thanks for the hint.For Ice -> water:

m = 0.1kg c=2100 J kg^-1 K^-1 dt = 10 degrees (-10 -> 0)

therefore E= mc dt = 0.1(2100)(10) = 2100 J

Energy required to melt the ice:

E= specific latent heat of fusion of ice * mass
= 325000(0.1)
= 32500 J

and then for water to steam:

m = 0.1 kg c= 4200 J kg^-1 K^-1 (for water) dt= 100 degrees (0->100)

=> E= 0.1 (4200) (100) = 42000 J

Energy required to vaporize 0.1 kg of water:

E = specific latent heat of vaporization of water * mass
= 225000(0.1)
= 22500 J

then i add them

2100 + 32500 + 42000 + 225000

and get 301600 J which is 301.6 kJ.Thank you.
 
That looks good.
Well done
Do you see that by far the greatest amount of energy is in converting liquid water to steam?
 
SORRY... misprint, you have the latent heat of vaporisation to convert water to steam out by factor of 10
Latent heat of vaporisation = 2300000 J/kg
I have only just noticed... fortunately you made a mistake in your multiplication so your answer is correct6.
should always double check calculations !
 
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