Check my answer for power output question

In summary, to determine the power output of an electric heater, a 0.25 kg piece of ice at -30°C was warmed and produced the following graph of temperature. Using the information on the graph, it was determined that the ice went from -30°C to -10°C in 150 seconds. Using the equation Q=mcΔt, the value of 3.3*105 was incorrectly used for the specific heat of ice, resulting in a calculation error. The correct calculation should be Q=0.25kg*2100J/(kg °c)*((-10°C)-(-30) °C) = 525*20°C = 10500J. Therefore, the power output of the electric heater
  • #1
MaryBarnes
27
0

Homework Statement



a 0.25 kg piece of ice at -30°C is warmed by an electric heater and the following graph of temperature is produced.
-use the information on the graph to determine the power output of the heater

the graph shows that the ice went from -30degC to -10 degC in 150s

Homework Equations


Q=mcΔt
p=w/Δt

The Attempt at a Solution


Q=mcΔt
=(0.25)(3.3*105)((-30)-(-20))
=82500*(-20)
1.65*106

should this be negative?

p=w/Δt
=1.65*106/150
=11000
 
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  • #2
You don't use units in your calculations, so they're hard to check.

When doing calculations on the rise in temperature of a substance, you want to subtract the colder temp. from the hotter temp.

Which is colder, -30 °C or -10 °C ?
(also, you have a typo in the temperatures used in your calculation of Q.

What is 3.3×105 ? This doesn't seem to be the heat capacity of ice.
 
  • #3
SteamKing said:
You don't use units in your calculations, so they're hard to check.

When doing calculations on the rise in temperature of a substance, you want to subtract the colder temp. from the hotter temp.

Which is colder, -30 °C or -10 °C ?
(also, you have a typo in the temperatures used in your calculation of Q.

What is 3.3×105 ? This doesn't seem to be the heat capacity of ice.

3.3*105 is the latent heat of fusion for water that I've been given, now i see where i may have messed up.
 
  • #4
MaryBarnes said:

Homework Statement



a 0.25 kg piece of ice at -30°C is warmed by an electric heater and the following graph of temperature is produced.
-use the information on the graph to determine the power output of the heater

the graph shows that the ice went from -30degC to -10 degC in 150s

Homework Equations


Q=mcΔt
p=w/Δt

The Attempt at a Solution


Q=mcΔt
=(0.25)(3.3*105)((-30)-(-20))
=82500*(-20)
1.65*106

should this be negative?
The value you've used for the specific heat of ice looks odd to me. Where did it come from and what are the units?
Note that the temperature change was positive, rising from -30C up to -20C. You've got the order of the temperatures reversed. You should have ΔT = Tfinal - Tinitial
p=w/Δt
=1.65*106/150
=11000
You might want to use different variable names for the change in time and change in temperature. It is common to use upper case T for temperatures and lower case t for times.

[Edit: I see that Steam King beat me to it. I'll leave it to him]
 
  • #5
im going to retry my answer. will post shortly
 
  • #6
=0.25kg*2100J/(kg °c)*((-10°C) - (-30) °C)
=525*20°C
=10500J

P=10500J/150s
=70w

is this ok?
 
Last edited:
  • #7
MaryBarnes said:
=0.25kg*2100J/(kg °c)*((-10°C) - (-30) °C)
=525*(-20)
=-10500J

P=10500J/150s
=70w

is this ok?

-10 - (-30) is how much? This is a simple problem in arithmetic.
 
  • #8
SteamKing said:
-10 - (-30) is how much? This is a simple problem in arithmetic.

its 20, but my answer has a negative. i was going to edit it but don't know how to
 
  • #9
MaryBarnes said:
its 20, but my answer has a negative. i was going to edit it but don't know how to
Just hit the edit button in the lower left hand corner.
 
  • #10
SteamKing said:
Just hit the edit button in the lower left hand corner.
I see now. Thanks!
 

1. What is power output?

Power output is a measure of the rate at which energy is transferred or converted from one form to another. It is typically measured in watts (W) or kilowatts (kW).

2. How do you calculate power output?

To calculate power output, you need to know the amount of energy transferred or converted and the time it took to do so. The formula for power output is: power output = energy transferred / time.

3. What is the difference between power output and power consumption?

Power output and power consumption are two related but distinct concepts. Power output refers to the amount of energy that is being transferred or converted, while power consumption refers to the amount of energy being used or consumed. Power output is typically measured in watts or kilowatts, while power consumption is typically measured in kilowatt-hours (kWh).

4. How does power output affect performance?

Power output is a key factor in determining the performance of a system or device. Generally, the higher the power output, the better the performance will be. This is because a higher power output means that energy is being transferred or converted at a faster rate, resulting in more work being done in a given amount of time.

5. Can power output be increased?

In most cases, it is possible to increase power output by making changes to the system or device. This can be achieved through various methods such as upgrading components, improving efficiency, or increasing the input of energy. However, there may be limitations depending on the specific system or device and its capabilities.

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