How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature?
2 answers:
<span>The amount of heat required to melt a substance at its melting point is given by
</span>
<span>
where m is the mass of the substance and Lf is its latent heat of fusion. During the phase change, the temperature of the substance does not change, and all the energy given to the substance is used to break the bonds between the molecules. In this problem, we have a block of ice of mass m=0.3 kg. The specific latent heat of ice is 334 kJ/kg: therefore, the amount of heat required to melt it is given by
</span>
<span>
So, the correct answer is A) 100375 J</span>
</span>
where m is the mass of the substance and Lf is its latent heat of fusion. During the phase change, the temperature of the substance does not change, and all the energy given to the substance is used to break the bonds between the molecules. In this problem, we have a block of ice of mass m=0.3 kg. The specific latent heat of ice is 334 kJ/kg: therefore, the amount of heat required to melt it is given by
</span>
So, the correct answer is A) 100375 J</span>
You might be interested in
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g