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Help me prove the units for m = kg

Hello

I've been going round in circles with this for a few days so thought some outside eyes might help :smile:

DSC_0448-1.jpg

(apologies for the rubbish photo!)

As you can see, I need to be left with only kg, but I can't see how to cancel the remaining units...can anyone nudge me in the right direction please?

Thanks!
Reply 1
The original equations are:

E = Pt and E = mc∆T + mL

Rearranged to make m the subject as follows:

Pt = mc∆T + mL

Pt = c∆T + L
m

m = 1
Pt c∆T + L

m = Pt
c∆T + L

Sorry they're not laid out the best, I don't know how to write equations on here!
Original post by *starbuck*
Hello

I've been going round in circles with this for a few days so thought some outside eyes might help :smile:

DSC_0448-1.jpg

(apologies for the rubbish photo!)

As you can see, I need to be left with only kg, but I can't see how to cancel the remaining units...can anyone nudge me in the right direction please?

Thanks!


Hey (:

I found a few mistakes in your answer.

In the third line, it should be kg m^2 s^-3 s at the top (you missed out the s^-1 that you multiplied to J)

At the bottom, because it's cT + L (and not cT x L), therefore the units of cT and the unit of L should be the same

i.e. the units at the bottom should be just kg m^2 s^-2 kg^-1.

And when you cancel both out, you get kg! (: Hope this helps.
In your final formula you have
m=Pt(cδT+L)m = \frac{Pt}{(c \delta T + L)}
The units on the right are Watt.second on the top.
Watt is Joule/second so the top is just Joule (J)
The bottom is
Jkg-1K-1.K + Jkg-1
This simplifies to
Jkg-1 + Jkg-1
which is just Jkg-1 on the bottom as you are just adding the same units.
(The units must be the same if you add two quantities.)
So you have
JJkg1\frac{J}{Jkg^-1}

Can you do the last step?
Reply 4
Ah thanks!!

I thought it might be something to do with the + on the bottom, but completely forgot how that was relevant. And yes! Missed a s-1 - I can totally see that now :smile:

Thanks both!! :smile:
just to inform the helpers here this is a TMA question from the open university and therefore you are helping someone to cheat

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