3200÷0.22= 145.4545...N
(it is an infinite decimal)
Answer:
For the first one c is the answer
For the second one c is also the answer
For the third one is b
Explanation:
I took that
The thermal efficiency is defined as follows
![\eta = 1 - \frac{Q_{\text{out}}}{Q_\text{in}}](https://tex.z-dn.net/?f=%5Ceta%20%3D%201%20-%20%5Cfrac%7BQ_%7B%5Ctext%7Bout%7D%7D%7D%7BQ_%5Ctext%7Bin%7D%7D)
,
and the energy which is put into the system is
![Q_{\text{in}} = W_\text{out} + Q_\text{out}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bin%7D%7D%20%3D%20W_%5Ctext%7Bout%7D%20%2B%20Q_%5Ctext%7Bout%7D)
.
In your case
![Q_\text{in} = 25 \text{ J},W_\text{out}=10 \text{ J}.](https://tex.z-dn.net/?f=Q_%5Ctext%7Bin%7D%20%3D%2025%20%5Ctext%7B%20J%7D%2CW_%5Ctext%7Bout%7D%3D10%20%5Ctext%7B%20J%7D.)
So
![Q_\text{out}=10 \text{ J}](https://tex.z-dn.net/?f=Q_%5Ctext%7Bout%7D%3D10%20%5Ctext%7B%20J%7D)
which gives an efficiency of
![\eta = 1 - \frac{10 \text{ J}}{25 \text { J}} = 0.6 = 60 \%](https://tex.z-dn.net/?f=%5Ceta%20%3D%201%20-%20%5Cfrac%7B10%20%5Ctext%7B%20J%7D%7D%7B25%20%5Ctext%20%7B%20J%7D%7D%20%3D%200.6%20%3D%2060%20%5C%25)
.
The question is incomplete, the complete question is;
A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answer:
See explanation
Explanation:
From Newton's law of cooling;
θ1 - θ2/t = K(θ1 + θ2/2 - θo]
Where;
θ1 and θ2 are initial and final temperatures
θo is the temperature of the surroundings
K is the constant
t is the time taken
Hence;
100 - 60/5 = K(100 + 60/2 - θo)
100 - 40/10 = K(100 + 40/2 - θo)
8= (80 - θo)K -----(1)
6= (70 - θo)K -----(2)
Diving (1) by (2)
8/6 = (80 - θo)/(70 - θo)
8(70 - θo) = 6(80 - θo)
560 - 8θo = 480 - θo
560 - 480 = -θo + 8θo
80 = 7θo
θo = 11.4°
Again from Newton's law of cooling;
θ = θo + Ce^-kt
Where;
t= 0, θ = 60° and θo = 11.4°
60 = 11.4 + C e^-K(0)
60 - 11.4 = C
C= 48.6°
To obtain K
40 = 11.4 + 48.6e^-10k
40 -11.4 = 48.6e^-10k
28.6/48.6 = e^-10k
0.5585 = e^-10k
-10k = ln0.5585
k= ln0.5585/-10
K= 0.0583
Hence, the temperature in 15 minutes;
θ= 11.4 + 48.6e^(-0.0583 × 15)
θ= 31.7°