Chemical Engineering - Chemical Engineering Basics - Discussion
Discussion Forum : Chemical Engineering Basics - Section 3 (Q.No. 15)
15.
The boiling & freezing points on a newly defined temperature scale in degree 'D' are 400°D & 100°D respectively. The temperature reading corresponding to 60°C on this new temperature scale will be equal to __________ °D.
Discussion:
9 comments Page 1 of 1.
Xyz said:
1 decade ago
Calibration of any temperature scale is done by some standard samples like WATER.
Here B.P. and M.P. of water is 100 and 0 degree Celsius respectively.
So 400*D = 100*C & 100*D = 0*C so for each 1*Celsius=3*D,
So 60*C = 100*D+(3*60)*D = 280.
Here B.P. and M.P. of water is 100 and 0 degree Celsius respectively.
So 400*D = 100*C & 100*D = 0*C so for each 1*Celsius=3*D,
So 60*C = 100*D+(3*60)*D = 280.
Shafeeq said:
9 years ago
100 * C = 400 * D.
60 * C = 400 x 60/100 = 240 * D.
60 * C = 400 x 60/100 = 240 * D.
BRT said:
9 years ago
240 is the right answer.
MOUNIKA said:
8 years ago
1*C = 4*D.
So 60*C = 60*4=240*D.
240 is right answer.
So 60*C = 60*4=240*D.
240 is right answer.
Ruwin said:
7 years ago
Check the accuracy of result by computing the equivalent of °D at 0°C.
1. 400°D = 100°C.
X°D. =. 0°C.
(400-100)°D/(100-0)°C=3°D/1°C.
General formula:
(400 - °D)/(100 - °C)=3/1.
(400 - °D) = 3x (100 - °C),
°D = 400 - 3 x (100 - °C),
So for 60°C
°D = 400 - 3 x (100 - 60),
°D = 400 - 3 x (40),
°D = 400 - 120,
°D = 280.
If 0°C
°D = 400 - 3 x (100 - 0),
°D = 400 - 3 x (100),
°D = 400 - 300,
°D = 100°D.
1. 400°D = 100°C.
X°D. =. 0°C.
(400-100)°D/(100-0)°C=3°D/1°C.
General formula:
(400 - °D)/(100 - °C)=3/1.
(400 - °D) = 3x (100 - °C),
°D = 400 - 3 x (100 - °C),
So for 60°C
°D = 400 - 3 x (100 - 60),
°D = 400 - 3 x (40),
°D = 400 - 120,
°D = 280.
If 0°C
°D = 400 - 3 x (100 - 0),
°D = 400 - 3 x (100),
°D = 400 - 300,
°D = 100°D.
Jitendra yadav said:
7 years ago
According to thermometric property.
Tn= a + bTc.
where Tn=N scale ; Tc=deg celcius scale.
Ice point : 100= a+b(0).(1).
Steam point : 400= a+b(100).(2).
Solving eqn. 1 & 2 , we get , a=100 b=3.
Corresponding to 60deg celcius , Tn= 100+3(60) =280 N [ans:D].
Tn= a + bTc.
where Tn=N scale ; Tc=deg celcius scale.
Ice point : 100= a+b(0).(1).
Steam point : 400= a+b(100).(2).
Solving eqn. 1 & 2 , we get , a=100 b=3.
Corresponding to 60deg celcius , Tn= 100+3(60) =280 N [ans:D].
(1)
Saurabh said:
5 years ago
The new scale is 100D --- 400D.
So the difference is 300D: 300D is the total measuring scope of the new scale.
60C= 100+300*60/100 (why added 100 because the new scale starting at 100D)
That's why
60C =100+ 180.
60C= 280D.
So the difference is 300D: 300D is the total measuring scope of the new scale.
60C= 100+300*60/100 (why added 100 because the new scale starting at 100D)
That's why
60C =100+ 180.
60C= 280D.
(1)
Mbhuyan said:
5 years ago
(D-100)/(400-100) = (60-0)/(100-0).
(D-100)/300 = 60/100,
D = 180 + 100,
= 280.
(D-100)/300 = 60/100,
D = 180 + 100,
= 280.
(2)
Ramsajan said:
3 years ago
100 + 60*3 = 280.
=> Because 300/100 = 3*60
=180 + 100 = 280.
=> Because 300/100 = 3*60
=180 + 100 = 280.
(1)
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