Question

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)

Answer 1

Step-by-step 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 * C * (T_(f) - T_(i))`

`6.51 kg * 9.8 * 66.8 m = 0.68 kg * 4200 J/kg^(o)C * (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`.