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melomori [17]
3 years ago
12

Suppose you walk into a sauna that has an ambient temperature of 48.0°C. (a) Calculate the rate of heat transfer in watts to you

by radiation given your skin temperature is 37.0°C, the emissivity of skin is 0.98, and the surface area of your body is 1.30 m2.
Physics
1 answer:
GenaCL600 [577]3 years ago
4 0

Answer:99.84 W

Explanation:

Given

Ambient temperature T_{\infty }=48^{\circ}C\approx 321\ K

Temperature of skin T=37^{\circ}C\approx 310\ K

Skin emissivity \epsilon =0.98

Surface area A=1.30\ m^2

Rate of heat transfer is Q=\epsilon A\sigma (T_0^4-T^4)

Q=0.98\times 1.30\times 5.67\times 10^{-8}((321)^4-(310)^4)

Q=99.84\ W

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If all else fails…

Explanation:

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You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

3 0
3 years ago
Carlos is making phosphorous trichloride using the equation below. He adds 15 g of phosphorus.
Degger [83]
Given:

The balanced chemical reaction of the synthesis of phosphorus trichloride:

2P + 3Cl2 ===> 2PCl3

Initial amount of phosphorus = 15 grams

The amount of product produced from 15 grams of phosphorus:

15 grams / 31 g/mol * (2/2) = 66.46 grams PCl3 

The amount of chlorine is 44.31 grams, nearest to 45 grams. 
8 0
3 years ago
Read 2 more answers
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ozzi

Answer:

the atomic mass is 11

Explanation:

the atomic mass is basically how many protons and neutrons there are so for this all you have to do is some simple math:

5 + 6 = 11

and boom, ur atomic mass is equal to 11!

4 0
2 years ago
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