# 1.6 Mechanisms of heat transfer  (Page 14/27)

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A man consumes 3000 kcal of food in one day, converting most of it to thermal energy to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?

2.59 kg

A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot is a 3.00-mm-thick callus with a conductivity at the low end of the range for wood and its density is $300\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$ . The area of contact is $25.0\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2},$ the temperature of the coals is $700\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , and the time in contact is 1.00 s. Ignore the evaporative cooling of sweat.

(a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a $1.40{\text{-m}}^{2}$ surface area? Assume that the animal’s skin temperature is $32.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , that the air temperature is $-5.00\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?

a. 39.7 W; b. 820 kcal

A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in $-1.00\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ water. The walrus’s internal core temperature is $37.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , and it has a surface area of $2.00\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ . What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood?

Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of $10.0\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ and a thermal conductivity twice that of glass wool with the rate of heat conduction through a 0.750-cm-thick window that has an area of $2.00\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ , assuming the same temperature difference across each.

$\frac{Q}{t}=\frac{kA\left({T}_{2}-{T}_{1}\right)}{d}$ , so that
$\frac{{\left(Q\text{/}t\right)}_{\text{wall}}}{{\left(Q\text{/}t\right)}_{\text{window}}}=\frac{{k}_{\text{wall}}{A}_{\text{wall}}{d}_{\text{window}}}{{k}_{\text{window}}{A}_{\text{window}}{d}_{\text{wall}}}=\frac{\left(2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.042\phantom{\rule{0.2em}{0ex}}\text{J/s}·\text{m}·\text{°}\text{C}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)\left(0.750\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\text{m}\right)}{\left(0.84\phantom{\rule{0.2em}{0ex}}\text{J/s}·\text{m}·\text{°}\text{C}\right)\left(2.00\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)\left(13.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\text{m}\right)}$
This gives 0.0288 wall: window, or 35:1 window: wall

Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is $1.40\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ ?

Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through the same area and at the same rate as computed in [link] , what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick.

$\frac{Q}{t}=\frac{kA\left({T}_{2}-{T}_{1}\right)}{d}=\frac{kA\text{Δ}T}{d}⇒$
$\text{Δ}T=\frac{d\left(Q\text{/}t\right)}{kA}=\frac{\left(6.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(2256\phantom{\rule{0.2em}{0ex}}\text{W}\right)}{\left(0.84\phantom{\rule{0.2em}{0ex}}\text{J/s}·\text{m}·\text{°}\text{C}\right)\left(1.54\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)}=1046\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}=1.05\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{K}$

One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose a single-story cubical house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, by what percentage would the heating cost of the house drop? Take the house to have dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors, and assume that the interior is uniformly at one temperature and the exterior is uniformly at another.

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