Question

Suppose you walk into a sauna that has an ambient temperature of 50.0
c. (a) calculate the rate of heat transfer to you by radiation in watts, given your skin temperature is 37.0 c, the emissivity of skin is 0.98, and the surface area of your body is 1.4 m2. (b) if all other forms of heat transfer are balanced (the net heat transfer is zero), how much will your body temperature increase if your mass is 70 kg, and sit inside the sauna for 1 h?

Answer 1

Hello mate which class question is this??

Answer 2

The change temperature is 1.885 K.

Explanation:

Given that,

Ambient temperature = 50.0°C

Skin temperature = 37.0°C

Emissivity of skin = 0.98

Area = 1.4 m²

Mass = 70 kg

time = 1 hour

(a). We need to calculate the rate of heat transfer

Using formula of rate of heat

$\dfrac{dQ}{dt}=\epsilon\times\sigma A(T_{a}^4-T_{s}^4)$

Where, $\epsilon$= emissivity

$T_{a}$=Ambient temperature

$T_{s}$= skin temperature

Put the value into the formula

$\dfrac{dQ}{dt}=0.98\times5.67\times10^{-8}\times1.4\times((50+273)^4-(37+273)^4)$

$\dfrac{dQ}{dt}=128.3\ watt$

(b). If all other forms of heat transfer are balanced

Heat transfer rate = 128.3 watt = 128.3 J/s

Heat transfer in 1 hour is

$Q=128.3\times3600\ J$

$Q=461880\ J$

We need to calculate the change temperature

Using formula of heat

$Q=mc\Delta T$

$\Delta T=\dfrac{Q}{mc}$

Put the value into the formula

$\Delta T=\dfrac{461880}{70\times3500}$

$\Delta T=1.885\ K$

Hence, The change temperature is 1.885 K.

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Topic : heat transfer rate

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