Respuesta :
Answer:
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.
[tex]ice(0\°C)\rightarrow water (20\°C)[/tex]
[tex]steam(1000\°C)\rightarrow water (20\°C)[/tex]
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 [tex]m_i[/tex] = 1.40 kg
Heat of fusion [tex]H_f[/tex] = 333 kJ/kg
Specific heat [tex]c[/tex] = 4186 J/kg°C = 4.186 kJ/kg°C
Heat of vaporization [tex]H_v[/tex] = 2260 kJ/kg
Let mass of steam required be = [tex]m_s[/tex] 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:
⇒ [tex]m_i\times H_f[/tex]
⇒ [tex]1.40\ kg\times 333 KJ/kg[/tex]
⇒ [tex]466.2\ kJ[/tex]
Heat required to raise the temperature to 20°C
⇒ [tex]m_i\times c\times (T_2-T_1)[/tex]
⇒ [tex]1.40\ kg\times 4.186 kJ/kg\°C\times (20\°C-0\°C)[/tex]
⇒ [tex]1.40\ kg\times 4.186 kJ/kg\°C\times 20\°C[/tex]
⇒ [tex]117.208\ kJ[/tex]
Heat required to condense steam to liquid at 100°C
⇒ [tex]m_s\times H_v[/tex]
⇒ [tex]m_s\ kg\times 2260 KJ/kg[/tex]
⇒ [tex]2260 m_s\ kJ[/tex]
Heat required to decrease temperature of steam to 20°C
⇒ [tex]m_s\times c\times (T_2-T_1)[/tex]
⇒ [tex]m_s\ kg\times 4.186 kJ/kg\°C\times (100\°C-20\°C)[/tex]
⇒ [tex]m_s\ kg\times 4.186 kJ/kg\°C\times 80\°C[/tex]
⇒ [tex]334.88 m_s\ kJ[/tex]
So, we can write as :
[tex]466.2+117.208=2260 m_s+334.88 m_s[/tex]
[tex]583.408=2594.88 m_s[/tex]
Dividing both sides by 2594.88
[tex]\frac{583.408}{2594.88}=\frac{2594.88 m_s}{2594.88}[/tex]
[tex]0.2248=m_s[/tex]
∴ [tex]m_s=0.2248\ kg\approx0.22\ kg[/tex]
Mass of steam required to raise the temperature of ice at 0°C to liquid water at 20°C is = 0.22 kg
