Question

A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wheel is immersed in 600 g of water at 15°C. What is the maximum temperature that the water could be increased to? Is this a very efficient way of heating water? The specific heat of water is 4.186 J/(g* ˚C).

Answer 1

Answer:
`16.234^(\circ)C`

Step-by-step explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water(
`m_w`)=600 gm

temperature of water =15

specific heat of water
`=4.186 j/g-^(\circ)C`

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.)
`=mgh=6* 9.81* 10=588.6 J`

Heat supplied
`=m_wc(\Delta T)`

`H.E.=600* 4.186* (T-16)`

`588.6=2511.6* (T-16)`

T-16=0.234

`T=16.234^(\circ)C`

This is not an efficient way of heating water as there is only
`0.234^(\circ)C`increase in temperature.