Question

The following data is obtained in a bomb calorimeter . Weight of crucible = 3.649 g ,Weight of crucible and fuel = 4.678 g ,Water equivalent of bomb calorimeter = 570 g ,Rise in temperature = 2.3 ˚C ,Fuse wire correction = 3.6 cal , Acid correction = 3.8 cal , Cooling correction =0.047 ˚C , Calculate gross and net calorific value if the fuel contain 6.5% hydrogen.​

Answer 1

Explanation:

Rearranging the data into ascending order, we get

54,57,59,60,62,62,62,65,70,75,84

Here,

(a) Mean = Sum of observations ÷ number of observations

=

12

54+54+57+59+60+62+62+62+65+70+75+84

=

12

788

=65.6 kg

(b) 4 people have a weight above the mean weight.

(c) Range = greatest observation − lowest observation

=84−54=30 kg

Answer 2

Answer:

The gross caloric value is equal to 6254.5 cal/gm and net calorific value if the fuel contain 6.5% hydrogen is ​5911.1 cal/gm.

Explanation:

Given that weight of crucible$= 3.649 g$

Weight of crucible and fuel $= 4.678 g$

Weight of fuel, m = 4.687 - 3.649 = 1.038g

Water equivalent of bomb calorimeter, w = 570 g

Water taken  in calorimeter, W = 2200g

Rise in temperature = t₂ - t₁ = 2.3 ˚C

Fuse wire correction , $C_F$ = 3.6 cal

Acid correction, $C_A$ = 3.8 cal

Cooling correction, $C_c$ =0.047 ˚C  

Cotton thread correction, $C_{CT}= 1.6\;cal$

Gross caloric value is the total quantity of heat liberated by one unit fuel is burnt and the products of the combustion has been cool to room temperature.

$G.C.V =\frac{(W+w)(t_2-t_1+C_c)-(C_A+C_F+C_{CT})}{m}$

$G.C.V= \frac{(2200+570)(2.3+0.047)-(3.8+3.6+1.6)}{1.038}$

G.C.V. = 6254.5 cal/gm

If the fuel contain 6.5% hydrogen, then net caloric value will be :-

$N.C.V. = HCV -$ latent heat of water vapor formed

N.C.V. = HCV - (0.09 × H×587)

N.C.V. = 6254.5 - (0.09 × 6.5 × 587)

N.C.V. = 6254.5 - 343.4

N.C.V. = 5911.1 cal/gm

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