Develop an expression for Temp as a function of time

[swmmr1928]

Homework Statement

A solid body at initial temperature T₀ is immersed in a bath of water at initial temperature T_w0. Heat is transferred from the solid to water at a rate \dot{Q}=K\bullet(T_w-T), where K is a constant and T_w and T are instantaneous values of the temperatures of the water and solid. Develop an expression for T as a function of time t. Check your result for the limiting cases, t=0 and t=∞. Ignore effects of expansion or contraction, and assume constant specific heats for both water and solid.

2. Homework Equations [/b[STRIKE]][/STRIKE]

\dot{Q}=K\bullet(T_w-T)

d(mU)_cv/dt=-\dot{Q}

C_v=dU/dT

The Attempt at a Solution

-K\bullet(T_w-T)=m*dU/dt

K\bullet(T-T_w)=m*C_v*dT/dt

dT/dt=K/(m*C_v)(T-T_w)

Now I will attempt integrating factor

dT/dt=c₁(T-T_w)

\mu(t)*dT/dt=\mu(t)*c₁(T-T_w)

\mu(t)=exp(∫-c₁*dt)=exp(-c₁*t)

d[T*exp(-c₁*t)]/dt=exp(-c₁*t)*c₁(T-T_w)

T*exp(-c₁*t)=∫exp(-c₁*t)*c₁(T-T_w)dt

T*exp(-c₁*t)=-exp(-c₁*t)(T-T_w)

T=-(T-T_w)

I would expect a the Temperature of the solid to decrease, as stated in the problem, but eventually level off at an asymptote as it approaches equilibrium.

 

[swmmr1928]

Maybe this should have been posted in math or introductory physics. can staff move it?

 

[swmmr1928]

As written, the equation is not separable. I attempted to use an integrating factor, but I am not sure if it is in the correct form. I obtained an equation that did not contain temperature at all, which is not what the question asks for.

 

[vela]

Homework Statement

A solid body at initial temperature T₀ is immersed in a bath of water at initial temperature T_w0. Heat is transferred from the solid to water at a rate \dot{Q}=K\bullet(T_w-T), where K is a constant and T_w and T are instantaneous values of the temperatures of the water and solid. Develop an expression for T as a function of time t. Check your result for the limiting cases, t=0 and t=∞. Ignore effects of expansion or contraction, and assume constant specific heats for both water and solid.

Homework Equations

\dot{Q}=K\bullet(T_w-T)

d(mU)_cv/dt=-\dot{Q}

C_v=dU/dT

Let's assume K>0. If heat moves from the solid to the water, you must have T>T_w, so $\dot{Q} = K(T_w-T) < 0$.

The Attempt at a Solution

-K\bullet(T_w-T)=m*dU/dt

K\bullet(T-T_w)=m*C_v*dT/dt

As the solid cools off, you want dU/dt < 0, so it looks like you have an extra negative sign. Your differential equation should be
\begin{align*}
m\frac{dU}{dt} &= K(T_w-T) \\
mc_v \frac{dT}{dt} &= K(T_w-T)
\end{align*}You're not ready to try solve this yet though because you don't know how T_w varies with time. You need to come up with another differential equation describing its behavior over time.

 

[swmmr1928]

\end{align*}You're not ready to try solve this yet though because you don't know how T_w varies with time. You need to come up with another differential equation describing its behavior over time.

\frac{dTw}{dt}=-\frac{dT}{dt}

 

[swmmr1928]

m*C_v*\frac{dT}{dt}=m'*C_v'*\frac{dTw}{dt}

 

[vela]

\frac{dTw}{dt}=-\frac{dT}{dt}

No, that's not correct. Try again.

 

[swmmr1928]

No, that's not correct. Try again.

Check my last post.

 

[swmmr1928]

Check my last post.

m*C_v*\frac{dT}{dt}=m'*C_v'*\frac{dTw}{dt}

 

[vela]

Oops, missed that. Looks okay except you're missing a minus sign.

 

[swmmr1928]

\frac{dTw}{dt}=\frac{m&#039;Cv&#039;}{K(Tw-T)}

 

[swmmr1928]

Yo, I'm not so good at this. What do I need to work on? Linear system of differential equations?

 

[vela]

\frac{dTw}{dt}=\frac{m&#039;Cv&#039;}{K(Tw-T)}

This is obviously wrong.

First, find an equation analogous to $mc_v \frac{dT}{dt} = K(T_w-T)$ for T_w. Once you have that, then you need to solve a system of linear differential equations.

Alternately, you could fix this equation
$$ mC_v\frac{dT}{dt} = m'C'_v\frac{dTw}{dt}, $$ integrate it, and use the result to eliminate T_w from your first differential equation.

In any case, it would help if you could explain what you're thinking. Right now, from your posts, it seems like you're just guessing. When you do that, people here aren't inclined to help as much.