What mass of steam at 100∘c must be added to 1.40 kg of ice at 0∘c to yield liquid water at 20 ∘c? the heat of fusion for water is 333 kj/kg , the specific heat is 4186 j/kg⋅c∘ , the heat of vaporization is 2260 kj/kg .

Mass of steam required to raise the temperature of ice at 0°C to liquid water at 20°C is  = 0.22 kg

Explanation:

Transformation occurring.

By law of conservation of energy, the heat transferred by steam to ice to melt it to liquid water is equal to the heat given out by steam to condense it to liquid water.

Given data:

Temperature of water = 20°C

Mass of ice = 1.40 kg

Heat of fusion = 333 kJ/kg

Specific heat   = 4186 J/kg°C = 4.186 kJ/kg°C

Heat of vaporization   = 2260 kJ/kg

Let mass of steam required be = kg

So, the equation can be given as:

Heat required to melt ice to liquid at 0°C+ Heat required to raise the temperature to 20°C = Heat  required to condense steam to liquid at 100°C + Heat required to decrease temperature to 20°C

Heat required to melt ice to liquid at 0°C:

⇒

⇒

⇒

Heat required to raise the temperature to 20°C

⇒

⇒

⇒

⇒

Heat  required to condense steam to liquid at 100°C

⇒

⇒

⇒

Heat required to decrease temperature of steam to 20°C

⇒

⇒

⇒

⇒

So, we can write as :

Dividing both sides by 2594.88

∴

Mass of steam required to raise the temperature of ice at 0°C to liquid water at 20°C is  = 0.22 kg

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer