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sdas [7]
3 years ago
8

How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Physics
1 answer:
Artyom0805 [142]3 years ago
7 0

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

             \text { 1 miles }=\frac{1}{0.621} \mathrm{km}

So, 2.50 miles will be equal to

            2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km.  So,

           1 \mathrm{km}=1000 \mathrm{m}

Thereby,

          4.026 \mathrm{km}=4026 \mathrm{m}

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Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

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  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
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<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

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= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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