Q. 1.60 - An Introduction to Thermal Physics

Question

A frying pan is quickly heated on the stovetop  ${200}^{\circ }\mathrm{C}$ It has an iron handle that is $20 cm$ long. Estimate how much time should pass before the end of the handle is too hot to grab with your bare hand. (Hint: The cross-sectional area of the handle doesn't matter. The density of iron is about$7.9\mathrm{g}/{\mathrm{cm}}^3$ and its specific heat is $0.45\mathrm{J}/\mathrm{g}{\cdot }^{\circ }\mathrm{C}$).

Step-by-Step Solution

Answer

From the following the time taken for the handle to heat is $8.25$ minutes

Step 1: Introduction

Assume we have a hot frying pan of  ${200}^{\circ }\mathrm{C}$ on a stove. Assuming that the iron handle is in good condition (length $20 cm$ ) the pan starts at room temperature ${20}^{\circ }\mathrm{C}$, The given density of iron is $7.9\times {10}^3\mathrm{kg}\cdot {\mathrm{m}}^{-3}$, its specific heat capacity is $c=450\mathrm{J}\cdot {\mathrm{kg}}^{-1}{\cdot }^{\circ }{\mathrm{K}}^{-1}$ and the thermal conductivity of iron is $k_t=80\mathrm{J}\cdot {\mathrm{s}}^{-1}{\cdot }^{\circ }{\mathrm{K}}^{-1}$, The rate of heat flow is:

$\frac{Q}{\mathrm{\Delta }t}=-k_{t}A\frac{\mathrm{\Delta }T}{\mathrm{\Delta }x}$

where $A$ is the cross-sectional area of the handle

Taking $\mathrm{\Delta }T=200-20={180}^{\circ }\mathrm{K}$ and $\mathrm{\Delta }x=0.2\mathrm{m}$, We can assume that all of the heat that passes along the handle is used to heat the iron (which isn't entirely true, because some of the heat will evaporate into the surrounding air, but we can ignore this because air's thermal conductivity is relatively low compared to iron's).

Step 2: Time taken by the handle to heat completely

The rate at which heat enters the handle is then measured.:

let it equation ($1$),

$\frac{Q}{\mathrm{\Delta }t}=80A\frac{180}{0.2}=7.2\times {10}^{4}A$

(Assuming a cylindrical handle), the mass of iron that is being heated is:

$m=\rho V=\rho hA=7.9\times {10}^3\times 0.2A$

$m=1580A\mathrm{kg}$ let be equation ($2$)

In time $\mathrm{\Delta }t$, therefore, the temperature will increase by (using equation ($1$)):

$\mathrm{\Delta }T=\frac{\left[Q\right]}{mc}=\frac{\left[7.2\times {10}^{4}A\mathrm{\Delta }t\right]}{mc}$

substitute from equation ($2$) and with $c=450\mathrm{J}\cdot {\mathrm{kg}}^{-1}{\cdot }^{\circ }{\mathrm{K}}^{-1}$, so:

$\mathrm{\Delta }T=\frac{\left[7.2\times {10}^{4}A\mathrm{\Delta }t\right]}{1580A\times 450}=0.101\mathrm{\Delta }t$ let be equation($3$)

A temperature of around $T_f={70}^{\circ }\mathrm{C}$ is probably hot, so we're looking for:

$\mathrm{\Delta }T=T_f-T_i$

Where, $T_i={20}^{\circ }$so:

$\mathrm{\Delta }T=70-20={50}^{\circ }\mathrm{K}$

So, from equation ($3$),

$\mathrm{\Delta }T=0.101\mathrm{\Delta }t\to \mathrm{\Delta }t=\frac{1}{0.101}\mathrm{\Delta }T=\frac{1}{0.101}\times 50$

$T=495\mathrm{s}=8.25\min$

So it would take around $8.25$ minutes for the handle to heat that much.

The mass of iron that is being heated (assuming a cylindrical handle) is: $\circ 70$ , therefore it would take around $8.25$ minutes for the handle to heat that much.