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Vladimir [108]
3 years ago
10

The specific heat of water is 4190 j/(kg*k). suppose you put 1 kg of water (a bit over 4 cups) into a microwave that can deliver

1000 w of power to the water when it is running at full power. roughly how long will you need to run the microwave at full power in order to raise the temperature of the water by 10°c?
Physics
1 answer:
VladimirAG [237]3 years ago
4 0
Given:
P = 1000 W, power input
c = 4190 J/(kg-K), the specific heat of water
m = 1 kg, mass of water
ΔT = 10 °C = 10 K, temperature rise.

Let t = time required to raise the temperature f the water.
Then
P*t = m*c*ΔT
(1000 J/s)*(t s) = (1 kg)*(4190 J/(kg-K)*(10 K)
1000t = 41900
t = 41.9 s

Answer: 41.9 s or 42 s (approximately)

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Answer:

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where G is Gibbs Free Energy

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          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

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∴ΔH = TΔS

  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

                                     = 163800 / 333

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This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

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                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

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