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2 years ago

a water heater has a power rating of 1 kW. how many seconds will this heater take to boil 1 liter of water?

1 answer:

5 0

Answer:
4.65 minutes.

Explanation:
Theoretical Assumptions (not realistic): 1) ALL of the heat energy is transferred to the water, and 2)the water is uniformly heated. 3) The water density is 1.0g/mL.
Because we are only heating it TO the “boiling point”, ONLY the energy to heat the water from 20 to 100’C using the heat capacity or specific heat of water is required. A “Watt” is 1 Joule/second. The specific heat of water is 4.184 J/g-’C. Water mass is 1000g. The energy required to heat it is therefore
1000
g
⋅
4.184
(
J
g
’
C
)
⋅
(
100
−
20
)
’
C
=
334720
J

The heater supplies 1200 J/s, so the time to provide this heat to the water is
334720
J
1200
(
J
s
)
=
278.9
s
or 4.65 minutes.

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2 years ago

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A man of mass m 1 5 70.0 kg is skating at v1 5 8.00 m/s behind his wife of mass m 2 5 50.0 kg, who is skating at v2 5 4.00 m/s.

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Answer:

A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)

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Say mass of husband is m1

Mass of the wife is m2

Velocity of the husband is v1

Velocity of the wife is v2

According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf

The momentum equation is

m1v1+m2v2= (m1+m2)vf

D. To solve for vf we need to make it subject of formula

vf= {(m1v1) +(m2v2)}/(m1+m2)

E. Substituting our given data

vf=

{(1570*58)+(2550*54)}/(1570+2558)

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Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

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where

speed at highest point = V2

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therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

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gh1 + $0.5V1^{2}$ = gh2

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now we can substitute all required values into the equation above.

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2 years ago

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