# Certain material heating water in a recipient

I don't know how to resolve a problem, but I don't want the answer since I'm almost going to have it resolved.

What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We want the water to be at 86ºC. The temperature of the iron pot is 12ºC. The water is heated by fire wood (65% of the thermal energy is wasted in combustion), and has a heating value of 12 MJ/kg. We need to determine the amount of wood needed to heat the water.

I would apply the normal formula of energy transfer: m1*c1*(t2-t1)=m2*c2*(t2-t1). But since we are using fire wood there, I don't know how to resolve this problem, probably because I'm missing something there... and I don't know what should I do.

Which formula or principle would I need to use when having the material and the heating value of the material used?

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(cont.) For example, what values do you have that you can plug into the equation you mentioned? Once you've figured that out, what values do you still need, and why aren't you able to get them? –  David Z Jan 7 '12 at 6:08
Hi David! Well, it's not actual homework but let's take it like that. I didn't posted the full problem since I didn't wanted an answer. What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We want the water to be at 86ºC. The temperature of the iron pot is 12ºC. The water is heated by fire wood (65% of the thermal energy is wasted in combustion), and has a heating value of 12 MJ/kg. We need to determine the amount of wood needed to heat the water. And I don't know how I should understand this with the formula I posted... –  Anne Jan 7 '12 at 6:12
OK, well, the homework tag is for homework-like questions so it probably applies even if this isn't actually homework. Also, we're not going to give you a full answer even if you do post the whole problem - it's right there in our homework FAQ that giving complete answers to homework-like questions is against the site policy. So don't worry about that! I'd suggest just editing the text of the problem (basically what you just put in your comment) into your question. –  David Z Jan 7 '12 at 6:14
... and it's done! Thanks David. I felt guilty last time because of that and didn't want to make the same mistake again. I'm trying to learn, that's why I don't want the final answer, but how to continue resolving my problem, since I'm unable to plug all the data in the formula - so probably I'm missing something as I said. –  Anne Jan 7 '12 at 6:18

If you know the specific heat of water and iron you can work out how much energy is needed to bring the water and the pot to 86°C. You need to work out how many kg of wood you need to burn to produce that amount of energy. The question even tells you how much of the energy produced by burning the wood is wasted and doesn't go towards heating the pot and water.

Whoever set the question is probably looking for the simple answer, but there are lots of extra bits you could have fun with. For example you're told the volume of the water not it's mass, but specific heat is normally quoted per unit mass. That means you need to know the density of water at 7°C to work out it's mass, and the density isn't exactly 1g/cm$^3$. Also the specific heat of both water and iron are functions of temperature.

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Which formula or principle would I need to use when having the material and the heating value of the material used?

I would suggest starting with the conservation of energy. Have you tried drawing a block diagram showing where the energy (heat) is going from/to within the limits of the setup?

Lastly keep an eye on the first law of thermodynamics...

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I would apply the normal formula of energy transfer: m1*c1*(t2-t1)=m2*c2*(t2-t1).

That's not the "normal formula of energy transfer". That's just the formula used when you mix liquids. A more generalized formula for energy transfer would be "heat lost=heat gained", or the first law of thermodynamics $$\Delta Q=\Delta U+\Delta W$$ Where $\Delta Q$ is heat supplied, $\Delta U$ is change in internal energy, and $\Delta W$ is work done by system.

In the case of your problem, as there is no work being done, just balance the heats. See how much heat goes into the system, and then compare this with the total heat required (sum of all $mc(T_2-T_1)$) to raise the temperature of the system.

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