Solving Physics Homework: Power Output of Heater

In summary: Once the liquid has reached its boiling point, the temperature stops increasing (as shown in the graph as a straight horizontal line), at this point, the additional heat only causes the particles to spread farther apart, until the liquid has completely boiled out, and has transformed into a gas. The temperature will then continue to rise.In summary, the power output of the heater is 70 watts, it will take 225 seconds to get the ice to the melting point of 0 degrees, and it will take 1171.4 seconds to melt the ice completely.
  • #1
alexandria
169
2

Homework Statement


upload_2016-4-7_12-40-20.png


Homework Equations


relevant equations listed with each question

The Attempt at a Solution


a)
to solve for the power output of the heater, i used the following equation: Power = Energy/time
i already know that Energy = Quantity
so based on this, i can solve for Q = mass(ice) x specific heat capacity (ice) x change in temperature (ice)
Q = (0.25 kg) x (2100 J/kg.C) x (0 degrees - (-30 degrees) = 15 750 J
Using Q, i substituted it into the equation to solve for power
P = 15 750 J / 150 s (150 s is the time shown on the graph)
P = 105 Watts
is this correct?
i need to know if this is right so i can use the power to do the next questions.

this is my second attempt to part a). the only difference is the change in temperature, instead of using 0 degrees, i used -10 degrees based on the graph above:
Q = (0.25 kg) x (2100 J/kg.C) x (-30 degrees - (-10 degrees) = 10 500 J
Using Q, i substituted it into the equation to solve for power
P = 10 500 J / 150 s
P = 70 Watts

i don't know if my attempts are correct, if anyone can clarify what I am supposed to do here, that would be greatly appreciated. Once i figure out what to do here, i can answer the next questions and post them on this forum for clarification as well.
 
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  • #2
alexandria said:
to solve for the power output of the heater, i used the following equation: Power = Energy/time
i already know that Energy = Quantity
so based on this, i can solve for Q = mass(ice) x specific heat capacity (ice) x change in temperature (ice)
Q = (0.25 kg) x (2100 J/kg.C) x (0 degrees - (-30 degrees) = 15 750 J
Using Q, i substituted it into the equation to solve for power
P = 15 750 J / 150 s (150 s is the time shown on the graph)
P = 105 Watts
is this correct?

The method is ok but check the graph for the temperature change that occurs in the 150 seconds (It's not 30C).
 
  • #3
the temperature change according to the graph is from -30 degrees which is the initial temperature, to -10 degrees. So considering this, is my second attempt correct then?
 
  • #4
Yes.
 
  • #5
ok so here are my final answers, please verify if they are correct. Thanks.
a)
Q = (0.25 kg) x (2100 J/kg.C) x (-30 degrees - (-10 degrees) = 10 500 J
Using Q, i substituted it into the equation to solve for power
P = 10 500 J / 150 s
P = 70 Watts
so the power output of the heater is 70 W

b)
time = Quantity / power
Q = (0.25 kg) x (2100 J/kg.C) x (0 degrees - (-30 degrees) = 15 750 J
time = 15750 J / 70 Watts
time = 225 s
it will take 225 seconds to get the ice to the melting point of 0 degrees.

c)
Q = mLf (Lf represents Latent heat of fusion)
Q = (0.25 kg) x (3.3 x 10^5 J/kg)
Q = 82 500 J
time = 82 500 J/70 watts
time = 1171.4 seconds

d) i don't need help with the graph

e) what does it mean when they say 'explain the graph according to the KMT"
this is my answer so far:
as the ice begins to increase in temperature, its particles begin to move faster and farther apart, and it reaches its melting point of 0 degrees. It requires a small amount of thermal energy to melt the ice, and that is shown in the diagram by the steep line. Once this happens, the temperature remains constant, and the additional heat only causes the particles to move farther apart, and increases the potential energy of the particles in the substance. That is why the graph shows a straight horizontal line when the ice is melting, because the temperature is constant and is not increasing. Once the ice completely melts and becomes a liquid, the temperature then continues to rise, more energy is required to turn this liquid into a gas. Once the liquid has reached its boiling point, the temperature stops increasing (as shown in the graph as a straight horizontal line), at this point, the additional heat only causes the particles to spread farther apart, until the liquid has completely boiled out, and has transformed into a gas. The temperature will then continue to rise.
 
  • #6
Your method looks right, though I haven't checked your arithmetic.
It requires a small amount of thermal energy to melt the ice, and that is shown in the diagram by the steep line. Once this happens, the temperature remains constant, and the additional heat only causes the particles to move farther apart, and increases the potential energy of the particles in the substance.
I believe you do understand what is happening, but your wording of this needs revision and improvement.

Bear in mind that energy involved in a phase change (e.g., from solid to liquid) is relatively huge.
 
  • #7
+1

One reason steam is so dangerous is that it can dump a lot of energy into you simply by condensing.
 
  • #8
so a) b) and c) are done correctly?
 
  • #9
Yes.
 
  • #10
srry for the late reply, but thanks for all the help everyone
i did the graph for part d on my own, here is the final result, is it accurate?
upload_2016-4-10_15-7-46.png
 

1. What is the formula for calculating power output of a heater?

The formula for calculating power output of a heater is P = IV, where P is power in watts, I is current in amperes, and V is voltage in volts. This equation is known as Joule's law.

2. How do I determine the current and voltage of a heater?

To determine the current and voltage of a heater, you can use a multimeter to measure the current and voltage at the source of the heater. Alternatively, you can refer to the specifications provided by the manufacturer.

3. Can I calculate power output without knowing the current or voltage?

No, you cannot calculate power output without knowing the current or voltage. These two values are necessary components of the formula for power output.

4. What units are used to measure power output?

Power output is measured in watts (W), which is equivalent to joules per second (J/s). You may also see power output measured in kilowatts (kW) or megawatts (MW) for larger power outputs.

5. How can I increase the power output of a heater?

To increase the power output of a heater, you can either increase the current or the voltage. However, it is important to note that increasing the power output may also increase the risk of overheating and damaging the heater.

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