91Ó°ÊÓ

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

$\frac{Q}{\mathrm{Δ}t}=−k_{t}A\frac{\mathrm{Δ}T}{\mathrm{Δ}x}$

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

Taking $\mathrm{Δ}T=200−20={180}^{∘}\mathrm{K}$and $\mathrm{Δ}x=0.2\mathrm{m}$, 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).

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

let it equation ($1$),

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

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

$m=\mathrm{ÒÏ}V=\mathrm{ÒÏ}hA=7.9×{10}^3×0.2A$

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

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

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

substitute from equation ($2$) and with $c=450\mathrm{J}â‹\dots {\mathrm{kg}}^{−1}{â‹\dots }^{∘}{\mathrm{K}}^{−1}$, so:

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

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

$\mathrm{Δ}T=T_f−T_i$

Where, $T_i={20}^{∘}$so:

$\mathrm{Δ}T=70−20={50}^{∘}\mathrm{K}$

So, from equation ($3$),

$\mathrm{Δ}T=0.101\mathrm{Δ}t→\mathrm{Δ}t=\frac{1}{0.101}\mathrm{Δ}T=\frac{1}{0.101}×50$

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

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: $∘70$, therefore it would take around $8.25$minutes for the handle to heat that much.