Radioactive decay of granite and other rocks in the Earth’s interior provides sufficient energy to
keep the interior molten, to heat lava and to provide warmth to hot springs. This is due to the
average release of about 0.03J per kilogram of granite each year. How many years are required
for a “chunk” of thermally insulated granite to increase in temperature by 400°C. The specific
heat capacity of granite is 800J/Kg°C.

Question

Radioactive decay of granite and other rocks in the Earth’s interior provides sufficient energy to
keep the interior molten, to heat lava and to provide warmth to hot springs. This is due to the
average release of about 0.03J per kilogram of granite each year. How many years are required
for a “chunk” of thermally insulated granite to increase in temperature by 400°C. The specific
heat capacity of granite is 800J/Kg°C.

anonymous · Answer

i actually figured out the answer with a bit of play with numbers.. but the funny thing is that I don't seem to understand it :S

shane_b · Answer

I think all you need is to use $$Q=mc \Delta t$$

anonymous · Answer

they kinda need time

anonymous · Answer

well, I multiplied c and (change in T) = 800 * 400... then divided that by the average release of 0.03 J/kg <--- i don't get what this bit of info means though

anonymous · Answer

the answer is 10666666.67 (time).. but what's the unit of time here?

shane_b · Answer

Shouldn't it just be seconds?

anonymous · Answer

i thought so at first, then I looked up some similar problems.. and it said years :S

shane_b · Answer

Yea, I'm getting that it's about 0.338 years

anonymous · Answer

where did that number come from ? :)

anonymous · Answer

seconds to years?

anonymous · Answer

yeah.. got that too

anonymous · Answer

what still appears foggy to me is that how come (J) divided by (J/kg) would give us time?

shane_b · Answer

Ugh...that should have been:

$$\frac{ 10666666.67s}{86400s/d * 365 d/yr}$$

shane_b · Answer

Yea, I'm not sure if that part is right...I was just looking at the value you came up with.  I need to think about the original problem for a few minutes...gotta make a phone call real quick though before the bank closes :)

anonymous · Answer

;) I'm here

shane_b · Answer

ok...hmm.  Let's see.

anonymous · Answer

:)

shane_b · Answer

I'm sort of playing around with the units to see what we should do here...

anonymous · Answer

I was trying to see if there's an correlation between Q and time.. but nothing so far

shane_b · Answer

$$Q=Joules=\frac{kg*m^2}{s^2}$$

anonymous · Answer

J/kg sould be:

anonymous · Answer

$$joules = N m => kg m/s^2    $$

anonymous · Answer

that per kg

anonymous · Answer

the thing is seconds are going to cancel out :S

shane_b · Answer

The m should be m^2 though

anonymous · Answer

sorry .. u're right, I forgot to multiply it by m

anonymous · Answer

the thing is, I know this question seems so simple... ugh.. physics is the only thing that makes me feel so naive..

shane_b · Answer

I know this is easy...we're just not "getting it" yet :)

anonymous · Answer

yep.

anonymous · Answer

could it be something that has to do with power = energy/time ?

anonymous · Answer

that's the only formula i know in physics that involves time and energy

shane_b · Answer

I thought about watts...but didn't think we needed it.  Maybe we do

anonymous · Answer

how can we figure out watts from the info given?

anonymous · Answer

we might not need it as a matter of fact.. take a look at this: http://en.wikipedia.org/wiki/SI_derived_unit

anonymous · Answer

around the bottom of the table, there's J/Kg converted to m^2/s^2 ... if we plug that to the formula of Q divided by J/kg ----> kgm^2/s^2 divided by m^2/s^2 = ...???

anonymous · Answer

I guess I might go back to it later.. but thanks for your help. :D

shane_b · Answer

Ok...It'll bother me until I figure it out..but I will look at it again later tonight :)

anonymous · Answer

^_^ great then.. might see you around

shane_b · Answer

ok, cya!