Physics, 23.12.2019 19:31 wambold3

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 .

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Answer from: tatenasira

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.

$ice(0\°C)\rightarrow water (20\°C)$

$steam(1000\°C)\rightarrow water (20\°C)$

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 m_i = 1.40 kg

Heat of fusion H_f = 333 kJ/kg

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

Heat of vaporization H_v = 2260 kJ/kg

Let mass of steam required be = m_s 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:

⇒ $m_i\times H_f$

⇒ $1.40\ kg\times 333 KJ/kg$

⇒ $466.2\ kJ$

Heat required to raise the temperature to 20°C

⇒ $m_i\times c\times (T_2-T_1)$

⇒ $1.40\ kg\times 4.186 kJ/kg\°C\times (20\°C-0\°C)$

⇒ $1.40\ kg\times 4.186 kJ/kg\°C\times 20\°C$

⇒ $117.208\ kJ$

Heat required to condense steam to liquid at 100°C

⇒ $m_s\times H_v$

⇒ $m_s\ kg\times 2260 KJ/kg$

⇒ $2260 m_s\ kJ$

Heat required to decrease temperature of steam to 20°C

⇒ $m_s\times c\times (T_2-T_1)$

⇒ $m_s\ kg\times 4.186 kJ/kg\°C\times (100\°C-20\°C)$

⇒ $m_s\ kg\times 4.186 kJ/kg\°C\times 80\°C$

⇒ $334.88 m_s\ kJ$

So, we can write as :

466.2+117.208=2260 m_s+334.88 m_s

583.408=2594.88 m_s

Dividing both sides by 2594.88

$\frac{583.408}{2594.88}=\frac{2594.88 m_s}{2594.88}$

0.2248=m_s

∴ $m_s=0.2248\ kg\approx0.22\ kg$

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

Answer from: Quest

where's the following? you gotta add it to know what the choices are

explanation:

Answer from: Quest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

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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?...

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