solution:
the change in the boiling point is given as,
dTbp =2.30°c
elevation constant for the solvent is given by,
kb=0.512°c/m

= 4.49m
Explanation:
The given data for case (1) is as follows.
h = 20 cm = 0.2 m
Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.
Q =
where, A = area
L = length
k = thermal conductivity = 0.8 W/m
= change in temperature.
Therefore, putting the given values into the above formula as follows.
Q =
=
= 168 W
For case (2), h = 180 cm = 1.8 m
Therefore, heat lost will be calculated as follows.
Q =
=
= 18.67 W
Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.
Answer:
John Dalton is the correct answer.
Explanation:
Answer:
1st answer: A. nitrogen
2nd answer: A. Sodium carbonate
Explanation:
Given that
Mass of water = 65.34 g
Amount of heat = mass of water * specific heat (temperature change
)
= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C
= 907.63 J
= 0.908 KJ
And
1 cal = 4.186798 J
907.63 J * 1 cal / 4.186798 J =216.78 cal
Or0.218 kcal