How to solve a problem about thermodynamics?

Physics
Tutor: None Selected Time limit: 1 Day

How much heat is needed in order to turn 10 grams of 0 degrees Celsius ice into 20 degrees Celsius water?

May 24th, 2015

Q = mc(t2 - t1)

c for water  = C = 4.186 joule/gram °C

Q = 10g*4.186 joule/gram °C*(20-10) = 418.6 Joules

May 24th, 2015

I don't understand what Q= mc(t2-t1) means.

May 24th, 2015

Q = Heat energy, if there was a temperature gained in a substance, this would be the heat energy "gained" in a substance.
If there was a temperature drop in a substance, this would be heat energy "lost" in a substance.

m = m for 'mass' as always, since you're trying to figure out how much heat energy that was applied, it will depend on the weight or in correct terms the "mass" of the the object.

c = basically it's a measure of how much heat energy the substance absorbs to change the temperature by 1 degree Fahrenheit/Celsius. (the unit for specific heat capacity is J kg^-1 K^-1)......(^ = to the power of)

Delta t = (t2 - t1): This can be the rise in temperature or the fall in temperature.

So, we have formula Q=mc*delta(t)

May 24th, 2015

wow, that makes so much sense thank you!

Can I confirm my answer with you?

May 24th, 2015

Yes, you can.

May 24th, 2015

I got two answers. One is 5,400J and 836J

May 24th, 2015

I' m sorry, I made a mistake in my previous solution. Solution below is correct.

Q = mc(t2 - t1)

c for water  = C = 4.186 joule/gram °C

Q = 10g*4.186 joule/gram °C*(20-0) = 837.2 Joules


May 24th, 2015

Q= 10g*540joule/gram= 5400 joule

Q= 10g*4.18joule/gram °C*(20°C)= 836 joule

This is how I solved it, is it correct? Also, I use 4.18 joule/gram °C not 4.186 joule/gram °C because that's what my teacher gave us, does it matter which one I used?

May 24th, 2015

Why do you write the first equation? It is not needed.

No, it does not matter what you use.

May 24th, 2015

I write the first equation because that's what I understood from the question.

How much heat is needed in order to turn 10 grams of 0°C ice into 20°C water? (Hint: you need to use two equations) c=4.18 J/g °C   Lf= 540 J/g

That is the complete question given

May 24th, 2015
Oh, it now makes sense: Q = mLf + mc*delta(t)
if you need to use this equation, then you've solved correctly.
But you need to add both equations to get final result:
Total Q = 5400 joule + 836 joule = 6236 joule
May 24th, 2015

Thank you so much!

May 24th, 2015

You are welcome:)

May 24th, 2015

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