Joules required to heat/how much heat is released

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To calculate the joules required to heat 250 grams of water from 25°C to 100°C, the formula q = mcΔT is applied, resulting in approximately 74,487 joules or 18,758 calories. For the heat released when 200 grams of steam condenses, the latent heat of water must be considered. The specific heat capacity of water is confirmed to be 4.186 J/g°C. The discussion also clarifies that ΔH in the equation q = ΔH*m refers to latent heat. Accurate calculations and understanding of these principles are essential for solving the problems presented.
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Homework Statement



1.)How many joules are required to heat 250 grams of liquid water from 25 C to 100 C

2.)How much heat is released when 200 grams of steam condenses?

Homework Equations



q=m*(delta) t*c
q=(delta)H*m

The Attempt at a Solution



1.)q=250g*75°C*4.186joules ? = 74 487joules ?
74 487 joules = 18,758 cal?
2.)
 
Last edited:
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jacob95 said:

Homework Statement



How many joules are required to heat 250 grams of liquid water from 25 C to 100 C

How much heat is released when 200 grams of steam condenses?

Homework Equations



q=m*(delta) t*c
q=(delta)H*m



The Attempt at a Solution



I don't know how to begin


For the first one, you have the formula

Q=mcΔT, so what is c for water ? You are given m and you can get ΔT (the change in temperature). So just plug in the numbers.

For the second one. What is the latent heat of water for this case of going from a gas to a liquid?
 
so would it be

q=250g*75°C*4.186j=78487.5 joules?
or
q=250g*75°C*1

then I would have to convert it to calories

78 487 joules = 18 758 cal?
 
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yes?
 
jacob95 said:
so would it be

q=250g*75°C*4.186j=78487.5 joules?
or
q=250g*75°C*1

then I would have to convert it to calories

78 487 joules = 18 758 cal?

It would be the first one since c= 4.186 J/g°C
 
thanks

what does ΔH stand for in

q=ΔH*m?
 
jacob95 said:
what does ΔH stand for in

q=ΔH*m?

I think that's the latent heat.
 
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