Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg
Answer:
Both Techs A and B
Explanation:
Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding and activates the anti lock braking system or traction system when it detects it is necessary.
Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.
Answer:
The statement can be written as
int result = cube(4);
Explanation:
A function is a block of reusable codes to perform some tasks. For example, the function in the question is to calculate the cube of a number.
A function can also operate on one or more input value (argument) and return a result. The <em>cube </em>function in the question accept one input value through its parameter <em>number </em>and the <em>number</em> will be multiplied by itself twice and return the result.
To call a function, just simply write the function name followed with parenthesis (e.g. <em>cube()</em>). Within the parenthesis, we can include zero or one or more than one values as argument(s) (e.g. <em>cube(4)</em>).
We can then use the "=" operator to assign the return output of the function to a variable (e.g. <em>int result = cube(4)</em>)
Answer:
Activation energy for creep in this temperature range is Q = 252.2 kJ/mol
Explanation:
To calculate the creep rate at a particular temperature
creep rate, ![\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )](https://tex.z-dn.net/?f=%5Czeta_%7B%5Ctheta%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7BR%20%5Ctheta%7D%20%29)
Creep rate at 800⁰C, ![\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )](https://tex.z-dn.net/?f=%5Czeta_%7B800%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7BR%20%28800%2B273%29%7D%20%29)
![\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\](https://tex.z-dn.net/?f=%5Czeta_%7B800%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7B1073R%7D%20%29%5C%5C%5Czeta_%7B800%7D%20%3D%201%20%5C%25%20per%20hour%20%3D0.01%5C%5C)
.........................(1)
Creep rate at 700⁰C
![\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )](https://tex.z-dn.net/?f=%5Czeta_%7B700%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7BR%20%28700%2B273%29%7D%20%29)
![\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}](https://tex.z-dn.net/?f=%5Czeta_%7B800%7D%20%3D%20C%20%5Cexp%28%5Cfrac%7B-Q%7D%7B973R%7D%20%29%5C%5C%5Czeta_%7B800%7D%20%3D%205.5%20%2A%2010%5E%7B-2%7D%20%20%5C%25%20per%20hour%20%3D5.5%20%2A%2010%5E%7B-4%7D)
.................(2)
Divide equation (1) by equation (2)
![\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%20%2A%2010%5E%7B-4%7D%20%7D%20%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20-%5Cfrac%7B-Q%7D%7B973R%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073R%7D%20%2B%5Cfrac%7BQ%7D%7B973R%7D%20%5D%5C%5CR%20%3D%208.314%5C%5C18.182%3D%20%5Cexp%5B%5Cfrac%7B-Q%7D%7B1073%2A8.314%7D%20%2B%5Cfrac%7BQ%7D%7B973%2A8.314%7D%20%5D%5C%5C18.182%3D%20%5Cexp%5B0.0000115%20Q%5D%5C%5C)
Take the natural log of both sides
![ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol](https://tex.z-dn.net/?f=ln%2018.182%3D%200.0000115Q%5C%5C2.9004%20%3D%200.0000115Q%5C%5CQ%20%3D%202.9004%2F0.0000115%5C%5CQ%20%3D%20252211.49%20J%2Fmol%5C%5CQ%20%3D%20252.2%20kJ%2Fmol)