How Fast Does Temperature Rise in an Unventilated Silicon Circuit?

[Pao44445]

Homework Statement

A silicon electric circuit is 23 mg, when electricity flows through the circuit rising the energy by 7.4 mW but this circuit wasn't design for heat ventilation. Find the rate of how much the temperature is rising per second ( ΔT/Δt ) The specific heat of silicon is 705 J/kg.K[/B]

Homework Equations

Q=mcΔT

The Attempt at a Solution

I know that when the electricity flows through the circuit, it will rise the temperature and we can figure the change in temperature by find both initial and final temperature but what about time? how this factor get in the calculation?

 

[DrClaude]

Can you express watts in terms of other SI units?

 

[Bystander]

Ask not for whom the bell tolls. Ask instead "the bell's heat capacity."

 

[Pao44445]

Can you express watts in terms of other SI units?

Joules per second ? it is just a rate of how much energy was transfer in 1 second, I can't plug this in Q :/

 

[DrClaude]

Joules per second ? it is just a rate of how much energy was transfer in 1 second, I can't plug this in Q :/

No, but you can calculate Q/Δt

 

[Pao44445]

No, but you can calculate Q/Δt

hmm I don't know if I am correct or not

from Q=mcΔT and ΔT/Δt
(Q/mc) / Δt = Q / (mc)Δt
7.4x10^-3 J / (23x10^-6kg)x(705 J/kg.K)x 1 s
= 0.46 K/s

 

[DrClaude]

Be careful with the mass there. Since you are given the heat capacity per gram, you can use g throughout and not convert to kg.

Also, I would find it more appropriate to take Q/Δt as a whole to be 7.4 mW instead of "assuming" 1 second.

 

[Pao44445]

Be careful with the mass there. Since you are given the heat capacity per gram, you can use g throughout and not convert to kg.

Also, I would find it more appropriate to take Q/Δt as a whole to be 7.4 mW instead of "assuming" 1 second.

edited
Then Q / (mc)Δt = W/mc
(7.4x10^-3 W ) / (23x10^-6kg)(705 J/kg.K)
= 0.46 K/s

but how this equation solve the problem?

 

[DrClaude]

Then Q / (mc)Δt = Wmc
(7.4x10^-3 W )(23x10^-6kg)(705 J/kg.K)
= 1.2x10^-4 W.J/K

but how this equation solve the problem?

You've missed a division in there.

 

[Pao44445]

You've missed a division in there.

oops, sorry. I've edited them and got the same answer 0.45-0.46 K/s
I think this is another pure mathematic problem :( really hate this kind of problem.

 

[DrClaude]

I wouldn't say that this is "pure mathematics." Actually, you need physical insight to figure out that you have information about Q/Δt. Once you figure that out, you have
$$
\frac{Q}{\Delta t} = m c \frac{\Delta T}{\Delta t}
$$
and you simply need to rearrange the equation to isolate ##{\Delta T}/{\Delta t}##, which is what you are asked for. Such simple algebra appears everywhere in physics, so you better get used to it (and good at it)

 

[Pao44445]

Never see that equation before :/

I don't know how come Q=mcΔT so I don't actually understand how to use it

 

[DrClaude]

Never see that equation before :/

I don't know how come Q=mcΔT so I don't actually understand how to use it

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html