Check my answer for power output question

[MaryBarnes]

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

 

[SteamKing]

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.

 

[MaryBarnes]

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.

 

[gneill]

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 = T_final - T_initial

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]

 

[MaryBarnes]

im going to retry my answer. will post shortly

 

[MaryBarnes]

=0.25kg*2100J/(kg °c)*((-10°C) - (-30) °C)
=525*20°C
=10500J

P=10500J/150s
=70w

is this ok?

 

[SteamKing]

=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.

 

[MaryBarnes]

-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

 

[SteamKing]

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.

 

[MaryBarnes]

Just hit the edit button in the lower left hand corner.

I see now. Thanks!