0.155 kg of water at 86.0oC is poured into an insulated cup containing 0.224 kg of ice initially at 0oC. How many kg of liquid will there be when the system reaches thermal equilibrium?
How much heat is lost by the water in going from 86 to zero. That will melt how much ice at zero?
To solve this problem, we need to consider the heat exchanged between the water and the ice in order to reach thermal equilibrium.
First, let's calculate the heat gained by the ice to reach the melting point:
Q_ice = m_ice * L_f
Where:
Q_ice is the heat gained by the ice
m_ice is the mass of ice
L_f is the latent heat of fusion of ice (which is 334 kJ/kg)
So, substituting the values into the equation, we have:
Q_ice = 0.224 kg * 334 kJ/kg = 74.816 kJ
Next, let's calculate the heat lost by the water to reach the melting point:
Q_water = m_water * c * (T_water_initial - T_melting)
Where:
Q_water is the heat lost by the water
m_water is the mass of water
c is the specific heat capacity of water (which is 4.186 kJ/kg°C)
T_water_initial is the initial temperature of water (86.0°C)
T_melting is the melting point of ice (0°C)
So, substituting the values into the equation, we have:
Q_water = 0.155 kg * 4.186 kJ/kg°C * (86.0°C - 0°C) = 56.256 kJ
Now, let's find the amount of heat needed to melt all the ice:
Q_melt = m_ice * L_f
Q_melt = 0.224 kg * 334 kJ/kg = 74.816 kJ
Since the system is insulated, the heat gained by the ice is equal to the heat lost by the water to melt the ice:
Q_ice = Q_water + Q_melt
74.816 kJ = 56.256 kJ + Q_melt
Q_melt = 74.816 kJ - 56.256 kJ = 18.560 kJ
Now, let's calculate the amount of liquid water formed after the ice melts:
m_water_final = Q_melt / c
m_water_final = 18.560 kJ / (4.186 kJ/kg°C) = 4.435 kg
Therefore, when the system reaches thermal equilibrium, there will be 4.435 kg of liquid water.