Hydrogen gas was cooled from 150c to 50c. its new volume is 75.0 ml. what was its original volume

1 answer:

6 0

From the information given, the hydrogen gas was cooled from a temperature of 150 °C to 50 °C. The final volume of the gas after being cooled came to 75ml. To find out its original volume, we need to apply Charles law which is expressed as : V1 /T1 = V2 /T2 Where V1 is original volume Where T1 is original temperature Where V2 is the final volume Where T2 is the final temperature. We need V1 so V1 = (V2 × T1) ÷ T2 V1 = (75 × 150) ÷ 50 V1 = 225 ml Original volume of the gas was 225 ml

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Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

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The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$