1.
How many calories are released by 50 grams of 10°C water when it freezes to 0°C ice?

Explanation:

Given that the masses of water is

m = 50grams

m = 50 / 1000 = 0.05 kg

Then, the water is raise from 10°C to ice of 0°C

As for the change in temperature, I will assume that it start off at room temperature 10°C

∆θ = 10-0 = 10

Then,

Specific heat capacity of water is

c = 4.186 J/g°C

Then, the amount calories can be calculated using the heat formula

H = mc∆θ

Where

c is the specific heat capacity of water

m is the mass of substance

∆θ is is change in temperature

H = mc∆θ

H = 50 × 4.186 × 10

H = 2093 J

The amount of calories released is 2093 J

4620Joules

Explanation:

Amount of heat used up will be the total calories of the water at 0°C expressed as:

H = mc∆t+mL

m is the mass of water

c is the specific heat capacity of water

∆t is the change in temperature

∆t = t2-t1

t2 is the final temperature

t1 is the initial temperature

Given m = 50grams = 0.05kg

c = 4200J/kg°C

∆t = 0°C - 10°C

∆t = -10°C

During change from 10°C to 0°C

H1 = mc∆t

H1 = 0.05(4200)(0-10)

H = 0.05×4200(-10)

H = -2100Joules

Heat energy gained by the ice at 0°C is expressed as

H2 = mLice

Lice is the latent heat of fusion of ice = 3.36 10^5 J Kg-1

H = 0.05 × 3.36 10^5

H2 = 6720Joules

Total calories released H = H1+H2

H = -2100+6720

H = 4620Joules

your answer is, orbit around the nucleus

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