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viva [34]
2 years ago
15

Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến

Physics
1 answer:
Mila [183]2 years ago
7 0

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

Power = P = \frac{E}{t}

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

P = \frac{2.927\ x\ 10^6\ J}{1200\ s}

<u>P = 2439.5 W = 2.439 KW</u>

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Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

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        Now equating this two formulas and making x the subject

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substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

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