- #1
kris24tf
- 35
- 0
Hello,
I have a couple problems that I have tried but I'm not sure if I worked them properly. Any advice or help is appreciated.
One problem is: An electric immersion heater has a power rating of 1500W. If the heater is placed in a liter of water at 20 degrees C, how many minutes will it take to bring the water to a boil? Assum no heat loss except to the water.
I tried it this way. Q-Pxt, P=1500W Q=mcDeltaT=1kg(4186)(-80 (change in temp))=334880. Then I took 334880 and divided by 1500 to get t... Is this the right way to do it or do I need to use a different equation?
The other isA volume of .5 L water at 16 deg. C is put into an aluminum ice tray of mass .25 kg at the same temp. How much energy must be removed from this system by the refrigerator to turn the water to ice at -8 deg. C?
I tried it by saying .5L=.5kg, Qgained=.25kg(920)(-8-16)=-5520
Qlost=.5(4186)(-24)=-50232, dividing these gives -.11 J... I am sure this is not the correct way. If someone could steer me in the right direction, I'd appreciate it...
Also, I do better with numerical explanations then written ones. What may sound normal to you might be over my head or sound completely different. Not trying to be weird, but it's true! And if anyone checks older posts, if there are any I haven't responded to in the past, it is only because I didn't get a response for a long time and did it on my own...
Please help me with these. I'd liek to know what I'm doing here...
I have a couple problems that I have tried but I'm not sure if I worked them properly. Any advice or help is appreciated.
One problem is: An electric immersion heater has a power rating of 1500W. If the heater is placed in a liter of water at 20 degrees C, how many minutes will it take to bring the water to a boil? Assum no heat loss except to the water.
I tried it this way. Q-Pxt, P=1500W Q=mcDeltaT=1kg(4186)(-80 (change in temp))=334880. Then I took 334880 and divided by 1500 to get t... Is this the right way to do it or do I need to use a different equation?
The other isA volume of .5 L water at 16 deg. C is put into an aluminum ice tray of mass .25 kg at the same temp. How much energy must be removed from this system by the refrigerator to turn the water to ice at -8 deg. C?
I tried it by saying .5L=.5kg, Qgained=.25kg(920)(-8-16)=-5520
Qlost=.5(4186)(-24)=-50232, dividing these gives -.11 J... I am sure this is not the correct way. If someone could steer me in the right direction, I'd appreciate it...
Also, I do better with numerical explanations then written ones. What may sound normal to you might be over my head or sound completely different. Not trying to be weird, but it's true! And if anyone checks older posts, if there are any I haven't responded to in the past, it is only because I didn't get a response for a long time and did it on my own...
Please help me with these. I'd liek to know what I'm doing here...