1. Jan 17, 2013

### chipM

1. The problem statement, all variables and given/known data

1)Wally whacko claims to have invented a heat engine that will revulutionalise the industry. It runs between a hot source at 300 C and a cold sink at 25 C . he claims that his engine is 92 % effieicnt.
What is the max efficiency of his engine ? in 2sf

(2)At a certain location the solar power per unit area reaching earth suface is 200 w/m^2. Averaged over 24 hour day.
If the average requirement in your home is 3.8 kw and you can convert solar power to elevtric power with 10% efficiency, how large a collector area will you need to meet all your house hold energy requirement from solar energy.

(3)The soecific heat capacity of ice is about 0.5 cal/g C.
Suppose that it remains at that value all the way to absolute zero. Calculate the number of calories it would take to change a 0.90g ice cube at absolute zero ( -273 C) t 0.90 g of boiling water.
How does this number of calories compare with the number of calories required to change the same gram of 100 C boiling water to 100 C steam. ?

2. Relevant equations
1) t(h)-t(c)/ t(h)
i dont know number 2
3) q=mct
3. The attempt at a solution

1) 300-25 /300 = 91.667 so its 91.667% efficeint
2) idont know where to start
3) o.90 x -273 x 0.5= -122.850

are they right ? please correct me if i am wrong with proper working

thanks !

2. Jan 17, 2013

### SammyS

Staff Emeritus
Hello chipM. Welcome to PF !

For #3: Why the negative sign when the change in temperature is positive?
How many calories does it take to melt ice -- latent heat of fusion.

What is the specific heat capacity of liquid water?​

3. Jan 17, 2013

### chipM

for liquid water its 4.18 i recon

thank you

4. Jan 17, 2013

### voko

For #1, use absolute temperature scale (kelvins).

For #2, start by finding how much solar power a solar panel has to consume to produce 3.8 kW at 10% efficiency.

For #3, ice goes from -273 to 0, melts, and then goes to +100.

5. Jan 17, 2013

### chipM

thanks mate

so i have to convert the temparature for 1) to kelvinss ?

so for 3) i got -173 x 0.09 x 0.5= -7.785 joules

correct ?

6. Jan 17, 2013

### SammyS

Staff Emeritus
Yes, or °Rankine if you prefer.

7. Jan 17, 2013

### chipM

cheers :)

so for 3) i got -173 x 0.09 x 0.5= -7.785 joules

correct ?

8. Jan 17, 2013

### voko

Not even close. As I said, the ice goes through three distinct stages on its way from abs. zero to boiling water.

9. Jan 17, 2013

### SammyS

Staff Emeritus
The temperature changes are positive. Why do you show it being negative again? This time, you seem to have ice going from -100°C to -273°C .

As voko pointed out:
For #3, ice goes from -273°C to 0°C, melts, and then goes to +100°C . ​

10. Jan 17, 2013

### chipM

k i am working on 3)

for 1) is it (573.15- 298.15)/ 573.15 = 0.48

which means its only 48% efficient ?

11. Jan 18, 2013

### SammyS

Staff Emeritus
Yes.

12. Jan 18, 2013

### chipM

hey man i still dont get how to do 2)

so is the temparature change 373 ?

which mean its going to be

0.5x 0.009x 373 = 0.168 .

yes ?

13. Jan 18, 2013

### SammyS

Staff Emeritus
You state that you don't know how to do 2, then jump right into #3 without mentioning that's what you're doing ????
No.

The temperature increase is 373°C, but the first 273 of that is for ice. (You do know that the ice melts at 0°C, right?)

How much heat does it take to melt 0.90 g of ice ?

Then the liquid water requires yet more heat to raise its temperature to the boiling point. (100°C at 1 atmosphere of pressure.) As I said before, the specific heat capacity of liquid water is different than that of ice. (Do you know what it is?)

After adding all that, how does that compare to the amount of heat required to change 0.90 g of liquid water to steam ?