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Lerok [7]
2 years ago
11

An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. What will be the final temperature of 500 g o

f water that is initially at 20ºC after 3 minutes if Does the water absorb all the heat given off by the heater? Does the water absorb 40% of the heat released?
Engineering
1 answer:
Lorico [155]2 years ago
8 0

Answer:

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An immersion heater has a resistance of 50Ω and carries a ...brainly.com › Engineering › High School

An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. ... What will be the final temperature of 500 g of water that is initially at 20ºC after 3 minutes if Does the water absorb all the heat given off by the heater? Does ...

Explanation:

Search Results

Web results

An immersion heater has a resistance of 50Ω and carries a ...brainly.com › Engineering › High School

An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. ... What will be the final temperature of 500 g of water that is initially at 20ºC after 3 minutes if Does the water absorb all the heat given off by the heater? Does ...

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Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power p
fgiga [73]

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

Net power output of the plant = 150kw

Solution:

The required collection rate

QH = W/n

= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.

= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

3 0
2 years ago
Find the altitude of the right cylinder of maximum convex surface that can be inscribed in a given sphere.
strojnjashka [21]

Answer:

The radius 4 is maximum in convex surface

5 0
2 years ago
A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac
WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

     = 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2

     =70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

    = (73.17 -53.3/53.3)*100

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5 0
3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
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