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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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Komok [63]

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Explanation:

Let's use Ohm's law:

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We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

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$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

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