# Heat required?

1. May 19, 2009

### Spaniard

1. How much heat is required to heat 5 kg. of water from 20C to steam at 100C?

I have no idea how to do this, i looked at some stuff online and i dont understand it, could someone explain how i can work throw this problem maybe even work throw it and explain the steps for me. Im more of a backwards learner if that makes sense. thanks

2. May 19, 2009

### Cyosis

What formulae does your text book provide with regards to this topic?

3. May 19, 2009

### minger

This is a fairly straightforward problem. All materials have what is called a specific heat capacity, which is quite literally the amount of heat to raise a unit mass a unit degree of temperature. The ubiquitous equation describing heat relationship to temperature is
$$Q = mC\Delta T$$
Where m is the mass, C is the specific heat capacity, and $$\Delta T$$ is the change in temperature.

This will only tell you the amount of heat needed to raise 5 kg of 20° water to 100° water though. There is additional heat required to go from 100° water to 100° steam.

Do you know any type of relation or property to that describes change in state?

4. May 19, 2009

### Spaniard

All i remmeber is it have a bunch of Q's and the heat was defined by KJ

5. May 19, 2009

### Spaniard

ok by online sources the formula to change water 100c to steam 100c is mLv, wat exactly is specific heat capacity? what i got so far is Q=5C80, would it help if i say im in a 100 level physics class?

6. May 19, 2009

### Cyosis

I have no idea what a physics level 100 class is. That said c is a material constant you will have to look it up in some kind of table book. The L is the specific latent heat which is the amount of energy you need to convert 1 kg of liquid to 1 kg of gas in this case. Again this is a material constant you will have to look it up. The total heat required for your problem is therefore the heat it takes to warm up water from 20 to 100 degrees plus the heat it takes to convert water at 100 degrees to steam at 100 degrees.

$$c_{water}=4.1813 kJ kg^{-1} K^{-1}, \;\;\; L_{water}=2260 kJ kg^{-1}$$

Do you not have a text book or any kind of information supplied by your physics class? These formulae should be contained within the text, including the numerical values for c and L.

Last edited: May 19, 2009
7. May 19, 2009

### Spaniard

ok so if i got this right $$Q = mC\Delta T$$ so Q=(5)(4.187)(80)=1674,8kj. So what does v stand for in mLv?? m is mass and L is specific heat constant of water right?

8. May 19, 2009

### Spaniard

im poor i dont got the text.... heh. and i cant barrow the text from a friend right now

9. May 19, 2009

### Cyosis

In the formula you listed v is just a subscript therefore $L_v$ is the specific latent heat. You have correctly calculated the energy needed to heat 5kg water from 20C to 100C. You also need to calculate the energy needed to convert all the water to steam.

10. May 19, 2009

### minger

How about class notes? If you are failing to grasp this concept, then things are going to be much, much more difficult for you very soon. I would either look into finding an International edition of the book, or working out some sort of arrangement such that you can have access to the class material.

11. May 19, 2009

### Spaniard

mv= 5 x 2257 = 1.30j. so over all it all equals 167.1J

12. May 19, 2009

### Cyosis

No that is not correct, 5*2257=11.3 kJ.