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3 years ago

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In a Joule experiment, a mass of 6.51 kg falls through a height of 66.8 m and rotates a paddle wheel that stirs 0.68 kg of water

. The water is initially at 15 degrees C. By how much does its temperature rise (in degrees C)

1 answer:

MatroZZZ [7]3 years ago

5 0

Explanation:

As the given data is as follows.

m = 6.51 kg, g = 9.8 $m/s^{2}$ , h = 66.8 m

mass of water (M) = 0.68 kg, Specific heat of water = 4200 $J/kg^{o}C$

$T_{1} = 15^{o}C$

According to the given situation, the decrease in potential energy of mass will be equal to heat energy gained by water.

Therefore,

mgh = $m \times C \times (T_{f} - T_{i})$

$6.51 kg \times 9.8 \times 66.8 m = 0.68 kg \times 4200 J/kg^{o}C \times (T_{f} - 15)^{o}C$

4261.7064 = $2856 (T_{f} - 15)^{o}C$

1.492 = $T_{f}$ - 15

$T_{f} = 16.492^{o}C$

So, $\Delta T = (16.492 - 15)^{o}C$

= $1.492^{o}C$

Therefore, we can conclude that rise in temperature will be $1.492^{o}C$ .

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Self-Inductance: -

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2 years ago

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We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

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For calculating number of turns

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Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

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$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

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$V_{2}=146.9 \mathrm{V}=V_{S}$

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$\frac{30}{N_{S}}=\frac{65}{146.9}$

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