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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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In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

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Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
3 years ago
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