# Heating water with carbon gas

1. Oct 10, 2014

### Karol

1. The problem statement, all variables and given/known data
200 liters of water are heated from 100 to 650C by burning coal gas. The heat loss in the chimney is 25%.
How many cubic meters of gas are needed

2. Relevant equations
Calories required: $Q=mc\delta t$
Burning heat of coal gas: 4320[Cal/m3]

3. The attempt at a solution
Net heat needed: $Q=200\cdot 55=11,000$
Gross heat needed: $Q\cdot 0.75=11,000\rightarrow Q=14,667$
Gas needed: $4320\cdot x=14,667\rightarrow x=3.4[m^3]$
It should be 2.9

2. Oct 10, 2014

### SteamKing

Staff Emeritus
You need to show units in all your calculations. It makes your work much easier to follow and analyze for mistakes.

3. Oct 10, 2014

### Karol

Net heat needed: $Q=200[kg]\cdot 55^0=11,000[Cal-kg]$
Gross heat needed: $Q[Cal-kg]\cdot 0.75=11,000[Cal-kg]\rightarrow Q=14,667[Cal-kg]$
Gas needed: $4320[Cal/m^3]\cdot x=14,667[Cal-kg]\rightarrow x=3.4[m^3]$
It should be 2.9[/QUOTE]

4. Oct 10, 2014

### nasu

You are missing something in the first formula. How can you get Calories by multiplying kg by degrees?

5. Oct 10, 2014

### Staff: Mentor

And how did the kg vanish again in the third line? This issue will go away once you fix the first line.

What should be 2.9?

6. Oct 10, 2014

### Karol

I mistakenly omitted the specific heat:
Net heat needed: $Q=200[kg]\cdot 1[\frac{Cal}{^0\cdot kg}]\cdot 55^0=11,000[Cal]$
Gross heat needed: $Q[Cal]\cdot 0.75=11,000[Cal]\rightarrow Q=14,667[Cal]$
Gas needed: $4320[Cal/m^3]\cdot x[m^3]=14,667[Cal]\rightarrow x=3.4[m^3]$

7. Oct 11, 2014

### NTW

Your answer is right, IMHO. Let's solve the problem by a false-position approach: we assume that you burn 2 cu. m of gas. The energy obtained is 4320*2 = 8640 Cal. You lose 25% through the chimney, and are left with 8640*0,75 = 6480 Cal. Now, you have 200 liters of water, so with those 6480 Cal, its temperature will rise by 6480/200 = 32,40ºC. But you need a rise of 55ºC, hence you need more gas... Exactly (55/32,4)*2 = 3,4 cu. m...

8. Oct 11, 2014

### Karol

Thanks

9. Oct 11, 2014

### Staff: Mentor

° is not a unit. Use K instead.
Apart from that, it looks fine (but I really don't like "Cal" for kcal).