When ice tends to absorb heat energy then it will change into liquid state. Now, change in the temperature of cold water $dT = T_{f} - 0^{o}C$ = $T_{f}$

Change in the temperature of original water in the container $dT' = 80 -T_{f
}$

Latent heat of melting of ice L = 80 cal/g

Specific heat of water C = 1 cal/g C

Heat absorbed by ice to melt = mass( $m_{i}$ ) x latent heat(L)

Heat absorbed by water = mass( $m_{w}$ ) x specific heat(C) x change in temperature (dT')

$m_{i}$ = 0.050 kg = 50 g and $m_{w}$ = 0.50 kg = 500 g

Now, heat lost by water is as follows.

heat lost by hot water( at $80^{o}C$ ) = heat absorbed by the ice + heat absorbed by cold water

$m_{i} \times L + m_{i} \times C dT = m_{w} \times C dT'$

$50 \times 80 + 50 \times 1 \times (T_{f}) = 500 \times 1 \times (80 - T_{f})$

So, we will cancel out 50 from each term given above. Hence,

$80 + T_{f} = 800 - 10T_{f}$

$11 T_{f}$ = 720

$T_{f} = 65.45^{o}C$

Thus, we can conclude that final temperature of the water is $65.45^{o}C$ .

6 0

3 years ago

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A submarine dives 185.0 m at an angle of 15.0◦ below the horizontal. What is the horizontal component of the submarine’s displac

stepan [7]

Answer:

Horizontal component, $d_x=178.69\ m$

Explanation:

It is given that,

Displacement of the submarine, d = 185 m

It is at an angle of 15 degrees below the horizontal. We need to find the horizontal component of the submarine's displacement. It is given by :

$d_x=d\ cos\theta$

Substituting the values in above equation as :

$d_x=185\times \ cos(15)$

$d_x=178.69\ m$

So, the horizontal component of the submarine’s displacement is 178.69 meters. Hence, this is the required solution.

3 0

3 years ago