Answer:
Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.
Explanation: lol
Answer:
Yes
Explanation:
Given Data
Temprature of source=750°c=1023k
Temprature of sink =0°c=273k
Work produced=3.3KW
Heat Rejected=4.4KW
Efficiency of heat engine(η)=
and
Heat Supplied 

η=
η=42.85%
Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink
η=1-
η=1-
η=73.31%
Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.
Answer:
Explanation:
First we compute the characteristic length and the Biot number to see if the lumped parameter
analysis is applicable.
Since the Biot number is less than 0.1, we can use the lumped parameter analysis. In such an
analysis, the time to reach a certain temperature is given by the following
From the data in the problem we can compute the parameter, b, and then compute the time for
the ratio (T – T)/(Ti
– T)
Answer:
#include <stdio.h>
void SplitIntoTensOnes(int* tensDigit, int* onesDigit, int DecVal){
*tensDigit = (DecVal / 10) % 10;
*onesDigit = DecVal % 10;
return;
}
int main(void) {
int tensPlace = 0;
int onesPlace = 0;
int userInt = 0;
userInt = 41;
SplitIntoTensOnes(&tensPlace, &onesPlace, userInt);
printf("tensPlace = %d, onesPlace = %d\n", tensPlace, onesPlace);
return 0;
}
Answer:
21.456 kJ/h
Explanation:
See the figure attached. In this case


Coefficient of performance in heat pump is defined by




Now it is necessary to change units, remember that Watt (W) is defined as J/s

