How Long Does It Take for a Handle to Become Unbearably Hot?

[WWCY]

Homework Statement

The question below is asking how long it would take for the cooler side of the handle to heat up till its unbearably hot.

I'm having a bit of trouble trying to understand the solution and would like some guidance.

I can't seem to get how the $\Delta T$ that represents the change in the cooler handle's average temperature is related to $\Delta T$ that came from $T - T_h$.

For example, if I had some spatial variable $x$, and $dx$ represents a small change in $x$, it makes sense to integrate over $xdx$.

In this case $\Delta T$ represents a function of the difference in temperature from end to end, and $d \Delta T$ doesn't seem like it's a small change of that function.

Could someone guide me towards the right interpretation? Thanks!

Homework Equations

The Attempt at a Solution

 

[DoItForYourself]

Hello,

You are right about this, ΔT is not the same in both sides of the equation.
In addition, T will be a function of x too, warmer near the second (hot) block and cooler in the other end.

But let's assume that T (in the handle block) is uniform and T is a function of time only (and not of distance). And let's assume that Δx=l/2 (the distance in which heat is transferred). Then the final equation is $$\frac {d(T-T_0)} {dt} = - \frac {T-T_h} {\tau}$$
where T₀ is the initial temperature of the handle block.

It seems that we can integrate this equation without problems because d(T-T₀)=dT=d(T-T_h) So, if we substitute d(T-T₀) with d(T-T_h) and separate the equation, we get $$\frac {d(T-T_h)} {T-T_h} = - \frac {dt} {\tau}$$ FInally, if we substitute T-T_h=ΔT, we get the same solution for ΔΤ (ΔT₀=T₀-T_h).

 

[WWCY]

Hello,

You are right about this, ΔT is not the same in both sides of the equation.
In addition, T will be a function of x too, warmer near the second (hot) block and cooler in the other end.

But let's assume that T (in the handle block) is uniform and T is a function of time only (and not of distance). And let's assume that Δx=l/2 (the distance in which heat is transferred). Then the final equation is $$\frac {d(T-T_0)} {dt} = - \frac {T-T_h} {\tau}$$
where T₀ is the initial temperature of the handle block.

It seems that we can integrate this equation without problems because d(T-T₀)=dT=d(T-T_h) So, if we substitute d(T-T₀) with d(T-T_h) and separate the equation, we get $$\frac {d(T-T_h)} {T-T_h} = - \frac {dt} {\tau}$$ FInally, if we substitute T-T_h=ΔT, we get the same solution for ΔΤ (ΔT₀=T₀-T_h).

Hi there, apologies for the delayed reply as I have been working on some other things recently.

Is this to say that the small increase in temperature of the cooler bit from its original temperature $d(T - T_0)$ is also a small increase of temperature towards $T_h$ as given by $\Delta T = T - T_h$?

Thanks for assisting!

 

[DoItForYourself]

Hi there, apologies for the delayed reply as I have been working on some other things recently.

Is this to say that the small increase in temperature of the cooler bit from its original temperature $d(T - T_0)$ is also a small increase of temperature towards $T_h$ as given by $\Delta T = T - T_h$?

Thanks for assisting!

Yes, because T₀ and T_h are constants.