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Ludmilka [50]
3 years ago
15

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa

ter is 4.186×10 to the third power J/kg times degrees Celsius
Physics
1 answer:
Anna [14]3 years ago
4 0

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature T_{1} = 68 °F = 20° c

Final temperature T_{2} = 212 °F = 100° c

Specific heat of water  C = 4.186 \frac{KJ}{kg c}

Now heat transfer Q = m × C × ( T_{2}  - T_{1} )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

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Answer:

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Answer:

Explanation:

1 )

Here

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δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

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For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

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