How Efficiently Can a 500 Watt Motor Heat Water?

[danago]

Hi. I have a set of questions i am supposed to do to get an idea about what to expect in an upcoming physics test. Ill post them, and show my working towards them.

A 500 watt electric motor is used to spin the blades of a mixer with a 50g container, containing 500g of water. If the motor is 35% efficient, how long will it take to increase the temperature of the water by 1 kelvin.

$$C_{container}=800 J kg^{-1} K^{-1}$$
$$C_{water}=4200 J kg^{-1} K^{-1}$$


For this question, since the motor is 35% efficient, i assumed that 65% of the energy it used was converted to heat energy. Since 500 J per second are being used by the motor, in 1 second, 325J will be converted to heat energy. To find how many seconds required to heat the water by 1K, i did the following:

$$Q=mc\Delta T$$
$$325t=0.5\times 4200\times 1$$
$$\therefore t=6.45 seconds$$

That could be a possible answer i guess, but i don't really understand how the container comes into it, or even if I've done it right.

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The temperature of water at the bottom of a waterfall will be higher than at the top, because of the gain in energy as it falls. If the gain in energy is given by:

$$Gain In Energy=mgh$$

where m is the mass of the water in kg, g is acceleration due to gravity in meters per second per second, and h is the distance the water falls in meters.

If a waterfall was 200m high, what would be the increase in temperature as the water falls?

$$g=9.8 m s^{-2}$$
$$C_w=4200 J kg^{-1} K^{-1}$$

This questions doesn't seem too hard. For this question, any value for m will give the same answer, so i said m=1 kg, thus, the gain in energy will equal 1960J, since g=9.8, and h=200. I then used the following equation and found how much the water would increase in temperature if it gained 1960J:

$$Q=mc\Delta T$$
$$1960=4200\times \Delta T$$
$$\therefore \Delta T=0.47$$

That is the answer what a few people who attempted the question in my class got, but according to the teacher, the answer is 4.7 degrees, and i have absolutely no clue how he got that; I am guessing its a mistake he has made.

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Why is heat loss greater through aluminium greater than that of timber window frames?

For this question, i just said that because aluminium has delocalised electrons while timber doesnt, it can conduct heat to a much better extent, so it will release heat to the atmosphere much quicker.

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How much water would have to evaporate per minute to take away all heat generated by the basic resting metabolizm of a 50kg person?

Assume the latent heat of vaporization is $22.5\times 10^5 J kg^{-1}$ and that the rate of metabolizm is $250 kJ h^{-1}$


This question i am completely lost. I have no idea what to do.

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1kg of a subctance is heated in a well insulated container by an electric heater with a power rating of 500 watts. The substance undergoes a temperature change through which it goes from a solid, to a liquid, to a gas. The heating curve for the substance is shown below.

http://img135.imageshack.us/img135/7051/heatcurve0ci.gif

What is the specific heat capacity of the substance in liquid form?


From the graph, i can see that while in liquid form, it rises 60 degrees before changing to a gas. That 60 degree rise takes place over 6 minutes, or 10 degrees per minute. In 1 second, the heater releases 500J of heat. So i could then say that 30kJ of heat is required to raise the temperature of the substance by 10 degrees? Since i want the heat capacity (energy per degree), i divide by 10, giving me 3kJ per degree, which is my answer.

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Thats all for now. I attempted what i could, and came to a reasonable answer, but I am not sure if theyre even close. Could anybody please read over my working and see if i have given a correct answer, and help me with questions that i have done wrong.

Thanks alot,
Dan.

 

[quark]

1. What portion of motor power is used for stirring. What happens to motor inefficient power (hint: where is the motor location?)

2. No issues.

3. Good

4. What is the quantity of heat to be removed from body? How much heat 1 kg of water can remove if it is evaporating? Doesn't it strike you like simple algebra?

5. Good.

 

[danago]

1. What portion of motor power is used for stirring. What happens to motor inefficient power (hint: where is the motor location?)

35% is used for stirring? 65% is inefficient power, which is converted to energies such as sound and heat (im assuming all heat in this case)? From there i don't know what to do

4. What is the quantity of heat to be removed from body? How much heat 1 kg of water can remove if it is evaporating? Doesn't it strike you like simple algebra?

Hmmm. Do i convert the rate of metabolizm to joules per minute? $4166.7J min^{-1}$. So every minute, i need to remove 4166.7J from the body via evaporation? If 1kg of water evaporated, $22.5\times 10^5J$ will be used. Is that right? I am not really sure where from here.

Thanks for the reply btw.

 

[Andrew Mason]

35% is used for stirring? 65% is inefficient power, which is converted to energies such as sound and heat (im assuming all heat in this case)? From there i don't know what to do

What is the means by which the motor generates heat in the water? Is it by the paddles or by the heat coming off the motor?

Hmmm. Do i convert the rate of metabolizm to joules per minute? $4166.7J min^{-1}$. So every minute, i need to remove 4166.7J from the body via evaporation? If 1kg of water evaporated, $22.5\times 10^5J$ will be used. Is that right? I am not really sure where from here.

So how much water turning to vapour each minute would consume 4166.7 Joules?

AM

 

[danago]

What is the means by which the motor generates heat in the water? Is it by the paddles or by the heat coming off the motor?

Is this a hint to me, or is it information i need to answer the question? because i thought about this, and wasnt really sure.

So how much water turning to vapour each minute would consume 4166.7 Joules?

AM

$$4166.7=m\times 22.5\times 10^5$$
$$\therefore m=0.00185kg=1.85g$$

Is that the answer?

 

[Hootenanny]

Is this a hint to me, or is it information i need to answer the question? because i thought about this, and wasnt really sure.

This was a hint. In otherwords, what heats up the water? The energy lost by the motor? Or the energy transferred to the water?

~H

 

[Andrew Mason]

Is this a hint to me, or is it information i need to answer the question? because i thought about this, and wasnt really sure.

How would the waste heat from the motor go into the water? Where does the mechanical energy from the motor (paddles) go? Why would you ignore it? Have a look at http://en.wikipedia.org/wiki/James_Prescott_Joule#The_mechanical_equivalent_of_heat" on mechanical energy and heat.

$$4166.7=m\times 22.5\times 10^5$$
$$\therefore m=0.00185kg=1.85g$$

Looks right to me.

AM

 

[danago]

Im just not understanding sorry. So are you saying that all of the waste energy from the motor is lost to the atmosphere, and none at all contributes to the heating of the water?

 

[Andrew Mason]

Im just not understanding sorry. So are you saying that all of the waste energy from the motor is lost to the atmosphere, and none at all contributes to the heating of the water?

I think you can assume that the waste heat from the motor does not go into the water. It goes into heating the motor and the air circulating through its coils. There is nothing in the problem that would allow you to determine how much, if any, of the heat might get into the water so assume it doesn't.

It is the mechanical work output of the motor that drives the paddles. The paddles churn the water and that energy is converted entirely into heat in the water (as Joule discovered).

AM

 

[danago]

ok, so if the motor is 35% efficient, 35% of its energy output will be mechanical, which then becomes heat. So in 1 second, 175J of heat goes towards heating the 500g of water.

$$Q=mc\Delta T$$
$$175t=0.5\times 4200\times 1$$
$$\therefore t=12$$

is that right?

 

[Andrew Mason]

ok, so if the motor is 35% efficient, 35% of its energy output will be mechanical, which then becomes heat. So in 1 second, 175J of heat goes towards heating the 500g of water.

$$Q=mc\Delta T$$
$$175t=0.5\times 4200\times 1$$
$$\therefore t=12$$

is that right?

For the water. But the container is heated as well because it is in direct contact with the water. Add the heat added to the container in raising the temperature of the container by one Kelvin.

AM