The relationships to determine the number of calories to change 0.50 kg of 0°C ice to 0°C ice water is 1,080,000 cal.
<h3>How does heating ice that is at C affect it?</h3>
Ice melts and becomes liquid water at 0 degrees Celsius. Once all of the ice has been entirely transformed into liquid water, the temperature of the remaining ice begins to increase once more (in °C), continuing to rise until it reaches 100 °C, where it then stabilizes.
The water turns into steam when it reaches a temperature of 100 °C (D).
Water has a fusion latent heat of fusion of 80 cal/g.
Water has a 1 cal/g-C specific heat.
Water has a 540 cal/g latent heat of vaporization.
In light of this, the total amount of heat needed is 1500 g [(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g [(720 cal/g)] = 1,080,000 cal.
To learn more about Vaporization refer to:
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Answer:
1/4
Explanation:
Mechanical Advantage = Load/Effort
Given
Effort applied = 24N
Load = 6N
Substitute
MA = 6/24
MA = 1/4
Hence the mechanical advantage is 1/4
Answer:
a ) 1.267 radian
b ) 1.084 10⁻³ mm
Explanation:
Distance of screen D = 1.65 m
Width of slit d = ?
Wave length of light λ = 687 nm.
Distance of second minimum fro centre y = 2.09 cm
Angle of diffraction = y / D
= 2.09 /1.65
= 1.267. radian
Angle of diffraction of second minimum
= 2 λ / d
so 2 λ / d = 1.267
d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm
=1084.45 nm = 1.084 x 10⁻³ mm.
