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3 years ago

How much heat energy in megajoules is needed to convert 7 kilograms of ice at -9c to water at 0c

Physics

2 answers:

gayaneshka [121]3 years ago

7 0

We can solve this by using the formulas provided to us by calorimetry.

The energy (heat) that is needed to change a given mass of some solid into its liquid form is

E=Lm·m (Where Lm is a specific constant for the material/substance and m is the mass)

Lm for solid water (i.e. ice) is 334 $\frac{Kj}{Kg}$

We can now solve for E

E= 334⋅2 Kj =668Kj

This is the minimum energy required to melt 2 kg of ice. Remember that this energy will in no way change the temperature of the water since every amount of this energy is needed just to melt the solid itself. Even more energy would be needed to increase the end temperature.

Musya8 [376]3 years ago

6 0

The latent heat of fusion of water is 344 j/p.This will take 344x300 or 100200 J to melt the ice. once it melted its at oc

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2 years ago

Two speakers, one directly behind the other, are each generating a 280-Hz sound wave. What is the smallest separation distance b

Annette [7]

Answer:

1.21m

Explanation:

If two speakers are generating a frequency of 280Hz, the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them is also known as the wavelength of the sound wave generated.

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Velocity v = frequency f × wavelength ¶

Given frequency = 280Hz, speed of sound v = 338m/s

Substituting this data's in the expression given to get the wavelength will give;

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3 years ago

What creates an ionic bond?

natta225 [31]

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3 years ago

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An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

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2 years ago