Heating Water with Coal Gas: Calculating Gas Consumption

[Karol]

Homework Statement

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

Homework Equations

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

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

 

[SteamKing]

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

 

[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]

 

[nasu]

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

 

[mfb]

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?

 

[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: $4320[Cal/m^3]\cdot x[m^3]=14,667[Cal]\rightarrow x=3.4[m^3]$
The answer should be 2.9[m³]

 

[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...

 

[Karol]

Thanks

 

[mfb]

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