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3 years ago

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A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh

eel 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).

1 answer:

maksim [4K]3 years ago

5 0

Answer: $16.234^{\circ}C$

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.) $=mgh=6\times 9.81\times 10=588.6 J$

Heat supplied $=m_wc(\Delta T)$

$H.E.=600\times 4.186\times (T-16)$

$588.6=2511.6\times (T-16)$

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.

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$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

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Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
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$Q = c \cdot m \cdot \Delta T$ ,

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