Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K
Explanation :
Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.
Formula used for efficiency of the heat engine.
![\eta =1-\frac{T_c}{T_h}](https://tex.z-dn.net/?f=%5Ceta%20%3D1-%5Cfrac%7BT_c%7D%7BT_h%7D)
where,
= efficiency = 0.780
= Temperature of hot reservoir = ?
= Temperature of cold reservoir = ![-24.8^oC=273+(-24.8)=248.2K](https://tex.z-dn.net/?f=-24.8%5EoC%3D273%2B%28-24.8%29%3D248.2K)
Now put all the given values in the above expression, we get:
![\eta =1-\frac{T_c}{T_h}](https://tex.z-dn.net/?f=%5Ceta%20%3D1-%5Cfrac%7BT_c%7D%7BT_h%7D)
![0.780=1-\frac{248.2K}{T_h}](https://tex.z-dn.net/?f=0.780%3D1-%5Cfrac%7B248.2K%7D%7BT_h%7D)
![T_h=1128.18K](https://tex.z-dn.net/?f=T_h%3D1128.18K)
Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K
Answer:
Yes
Explanation:
You are using energy to click the mouse, and the energy moves from your fingers to the mouse clicker.
Answer:
The magnetic field in the System is 0.095T
Explanation:
To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.
By Faraday's law we know that
![\epsilon = \frac{NBA}{t}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20%5Cfrac%7BNBA%7D%7Bt%7D)
Where,
electromotive force
N = Number of loops
B = Magnetic field
A = Area
t= Time
For Ohm's law we now that,
V = IR
Where,
I = Current
R = Resistance
V = Voltage (Same that the electromotive force at this case)
In this system we have that the resistance in series of coil and charge measuring device is given by,
![R = R_c + R_d](https://tex.z-dn.net/?f=R%20%3D%20R_c%20%2B%20R_d)
And that the current can be expressed as function of charge and time, then
![I = \frac{q}{t}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7Bq%7D%7Bt%7D)
Equation Faraday's law and Ohm's law we have,
![V = \epsilon](https://tex.z-dn.net/?f=V%20%3D%20%5Cepsilon)
![IR = \frac{NBA}{t}](https://tex.z-dn.net/?f=IR%20%3D%20%5Cfrac%7BNBA%7D%7Bt%7D)
![(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bq%7D%7Bt%7D%29%28R_c%2BR_d%29%20%3D%20%5Cfrac%7BNBA%7D%7Bt%7D)
Re-arrange for Magnetic Field B, we have
![B = \frac{q(R_c+R_d)}{NA}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7Bq%28R_c%2BR_d%29%7D%7BNA%7D)
Our values are given as,
![R_c = 58.7\Omega](https://tex.z-dn.net/?f=R_c%20%3D%2058.7%5COmega)
![R_d = 45.5\Omega](https://tex.z-dn.net/?f=R_d%20%3D%2045.5%5COmega)
![N = 120](https://tex.z-dn.net/?f=N%20%3D%20120)
![q = 3.53*10^{-5}C](https://tex.z-dn.net/?f=q%20%3D%203.53%2A10%5E%7B-5%7DC)
![A = 3.21cm^2 = 3.21*10^{-4}m^2](https://tex.z-dn.net/?f=A%20%3D%203.21cm%5E2%20%3D%203.21%2A10%5E%7B-4%7Dm%5E2)
Replacing,
![B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%283.53%2A10%5E%7B-5%7D%29%2858.7%2B45.5%29%7D%7B120%2A3.21%2A10%5E%7B-4%7D%7D)
![B = 0.095T](https://tex.z-dn.net/?f=B%20%3D%200.095T)
Therefore the magnetic field in the System is 0.095T
Charles Law
Explanation:
Step 1:
It is given that the original volume of the gas is 250 ml at 300 K temperature and 1 atmosphere pressure. We need to find the volume of the same gas when the temperature is 350 K and 1 atmosphere pressure.
Step 2:
We observe that the gas pressure is the same in both the cases while the temperature is different. So we need a law that explains the volume change of a gas when temperature is changed, without any change to the pressure.
Step 3:
Charles law provides the relationship between the gas volume and temperature, at a given pressure
Step 4:
Hence we conclude that Charles law can be used.