Question 1/5
1 answer:
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Answer:
t = 6179.1 s = 102.9 min = 1.7 h
Explanation:
The energy provided by the resistance heater must be equal to the energy required to boil the water:
E = ΔQ
ηPt = mH
where.
η = efficiency = 84.5 % = 0.845
P = Power = 2.61 KW = 2610 W
t = time = ?
m = mass of water = 6.03 kg
H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)
<u>t = 6179.1 s = 102.9 min = 1.7 h</u>
Answer:
Bore = 7 cm
stroke = 6.36 cm
compression ratio = 10.007
Explanation:
Given data:
Cubic capacity of the engine, V = 245 cc
Clearance volume, v = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
V =
or
V =
on substituting the values, we have
245 =
or
D = 7.00 cm
Now,
we have
D/L = 1.1
thus,
L = D/1.1
L = 7/1.1
or
L= 6.36 cm
Now,
the compression ratio is given as:
on substituting the values, we get
or
Compression ratio = 10.007