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## Homework Statement

This is a lab report that I'm working on. I'm almost done, except for the last part which consists of finding the error in the result for our heat transfer coefficient h using Newton's law of cooling. I've tried this in different ways but I can't seem to get it right.

We have 51 data points for temperature and 51 data points for their respective times.

## Homework Equations

ln( (T-T(room))/(T(initial) - T(room)) = -hAt/(C*p*V)

Where T is the variable of temperature of the metal sphere whose temperature is being measured, T(room) is room temperature, T(initial) is initial temperature of sphere, h is the heat transfer coefficient, A is the surface area, t is the time, C is the specific heat, p is the density and V is the volume.

The root sum square method says, that if we have h(T,t), then the uncertainty of h:

Uh = +/- [ (T/h * ∂h/∂T * U_T)^2 + (t/h * ∂h/∂t * U_t)^2 ] ^(1/2)

U_T is the uncertainty of the measurement device, which is 0.5 degrees C. U_t I assume will be 1 second, since the timer was read to the nearest second.

## The Attempt at a Solution

First, solving for h(T,t):

h = - [CpVln((T-T(room))/(T(initial)- T(room)) ]/(At)

Partial derivatives:

∂h/∂T = -CpV/(A*t*(T-T(room)))

∂h/∂t = [CpVln((T-T(room))/(T(initial)- T(room)) ]/(At^2)

Multiplying by T/h and t/h for the respective components above:

∂h/∂T *T/h = 1/(T-T(room)) * (1/ln((T-T(room))/(T(initial)- T(room))

∂h/∂t * t/h = -1

But then when placing these two in equation for U_h, I'm not sure which T to use for the portion of ∂h/∂T *T/h*U_t.

Another way that I tried solving this was by letting f(T) = ln(T-T(room)/(Tin - T(room)).

But then the function will be come h= h(f(T), t), so the uncertainty would be:

U_h = +-[ (∂h/∂f(T) *f(T)/h*U_f(T))^2 + (1)^2] ^(1/2).

∂h/∂f(T) * f(T)/h would reduce down to -1, so I would get:

U_h = +-[ (-U_f(T))^2 +1 ]^(1/2).

I wasn't sure if this second method was right, and if it is, what would I use for the uncertainty of the f(T) function U_f(T)?

I hope this post makes sense. I've been working on this particular part of the lab for so long. Any help would be greatly appreciated.