A common cause of thermal pollution is the use of water as a coolant by power plants and industrial manufacturers
Answer:
the final temperature is T=305.63 K
Explanation:
using the Stephan-Boltzmann equation for black bodies
q = σ*(T⁴-T₀⁴)
where
q= heat flux = 155 W/m²+150 W/m² = 255 W/m²
σ= Stephan-Boltzmann constant = 5.67*10⁻⁸ W/m²K⁴
T= absolute temperature
T₀= absolute initial temperature = 255 K
solving for T
q = σ*(T⁴-T₀⁴)
T = (q/σ + T₀⁴)^(1/4)
replacing values
T = (q/σ + T₀⁴)^(1/4) = (255 W/m²/(5.67*10⁻⁸ W/m²K⁴) + (255 K)⁴)^(1/4) = 305.63 K
T=305.63 K
thus the final temperature is T=305.63 K
2/3 is the fraction of the carbon dioxide exhaled by all of the animals that is generated by the reactions of the citric acid cycle, if the gluscose is the sole energy source. So if the glucose is the energy source the fraction of the carbon dioxide is 2/3. The answer in this question is 2/3
Answer:
To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law. The equation follows:
\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}
T
1
P
1
V
1
=
T
2
P
2
V
2
where,
P_1,V_1\text{ and }T_1P
1
,V
1
and T
1
are the initial pressure, volume and temperature of the gas
P_2,V_2\text{ and }T_2P
2
,V
2
and T
2
are the final pressure, volume and temperature of the gas
We are given:
\begin{gathered}P_1=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K\end{gathered}
P
1
=760mmHg
V
1
=175L
T
1
=15
o
C=[15+273]K=288K
P
2
=640mmHg
V
2
=198L
T
2
=?K
Putting values in above equation, we get:
\begin{gathered}\frac{760mmHg\times 175L}{288K}=\frac{640mmHg\times 198L}{T_2}\\\\T_2=274K\end{gathered}
288K
760mmHg×175L
=
T
2
640mmHg×198L
T
2
=274K
Hence, the temperature when the volume and pressure has changed is 274 K
