A 206.3 kg walrus initially at rest on a horizontal floor requires a 249 N horizontal force to set it in motion. Find the coeffi
1 answer:
You might be interested in
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>:
- mass of ice, m= 0.555 kg
- temperature of ice, T= -16.6°C
- rate of heat transfer,
- specific heat of ice,
- latent heat of fusion of ice,
<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>
Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.
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.
= 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,
<u>Now the time required for supply of 185370 J:</u>
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
The elastic potential energy (Ep) is given by 
Data:
Ep = ? (Joule)
k = 20 N/m
x (displacement) = 0.20 m
Solving:
Data:
Ep = ? (Joule)
k = 20 N/m
x (displacement) = 0.20 m
Solving: