# Thermal energy

1. Apr 7, 2016

### alexandria

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

2. Relevant equations
relevant equations listed with each question
3. 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 dont know if my attempts are correct, if anyone can clarify what im 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.

2. Apr 7, 2016

### CWatters

The method is ok but check the graph for the temperature change that occurs in the 150 seconds (It's not 30C).

3. Apr 7, 2016

### alexandria

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. Apr 7, 2016

Yes.

5. Apr 7, 2016

### alexandria

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 dont 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. Apr 7, 2016

### Staff: Mentor

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. Apr 8, 2016

### CWatters

+1

One reason steam is so dangerous is that it can dump a lot of energy into you simply by condensing.

8. Apr 8, 2016

### alexandria

so a) b) and c) are done correctly?

9. Apr 8, 2016

### CWatters

Yes.

10. Apr 10, 2016

### alexandria

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?