Joules required to heat/how much heat is released

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Homework Help Overview

The discussion revolves around calculating the energy required to heat water and the heat released during the condensation of steam. The subject area includes thermodynamics and heat transfer concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the formula q=m*(delta) t*c for heating water and question the values for specific heat and latent heat. There are attempts to calculate the required joules and conversions to calories, along with inquiries about the meaning of ΔH in the context of the equations.

Discussion Status

The discussion includes various attempts to apply the relevant equations, with some participants expressing uncertainty about the initial steps and the values needed. Guidance has been offered regarding the specific heat of water and the latent heat concept, but no consensus has been reached on the calculations.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a focus on understanding the definitions and values associated with the equations 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?
 
Last edited:
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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