Inside an insulated container, a 250-g ice cube
at 0°C is added to 200 g of water at 18°C.
(i) What is the final temperature of the system?
(ii) What is the remaining ice mass?
To melt all of the ice, 250g*80cal/g = 20,000 calories must be added. The already-liquid water can transfer only 200*18 = 3600 calories as it cools from 18 to 0 C. Therefore some of the ice remains at equilibrium. (i) The final temperature is 0 C.
(ii) Bringing all of the original liquid from 18C to 0C transfers 3600 cal to the ice. That is enough to melt 3600/80 = 45 g.
205 g of ice remains.
To find the final temperature of the system, we can use the principle of energy conservation. The heat lost by the water will equal the heat gained by the ice cube.
We can calculate the heat lost by the water (Q_lost_water) using the formula:
Q_lost_water = m_water * C_water * (T_final - T_initial_water)
where:
m_water = mass of water (200 g)
C_water = specific heat capacity of water (4.18 J/g°C, approximately)
T_final = final temperature of the system (what we're trying to find)
T_initial_water = initial temperature of the water (18°C)
Similarly, we can calculate the heat gained by the ice cube (Q_gained_ice) using the formula:
Q_gained_ice = m_ice * L_f + m_ice * C_ice * (T_final - T_initial_ice)
where:
m_ice = mass of ice cube (250 g)
L_f = latent heat of fusion of ice (334 J/g, approximately)
C_ice = specific heat capacity of ice (2.09 J/g°C, approximately)
T_initial_ice = initial temperature of the ice cube (0°C)
Since the system is insulated, there is no heat exchange with the surroundings, and the heat lost by the water will be equal to the heat gained by the ice cube, so:
Q_lost_water = Q_gained_ice
Substituting the values into the equations:
m_water * C_water * (T_final - T_initial_water) = m_ice * L_f + m_ice * C_ice * (T_final - T_initial_ice)
Simplifying the equation:
m_water * C_water * T_final - m_water * C_water * T_initial_water = m_ice * L_f + m_ice * C_ice * T_final - m_ice * C_ice * T_initial_ice
Rearranging the equation to solve for T_final:
m_water * C_water * T_final - m_ice * C_ice * T_final = m_water * C_water * T_initial_water - m_ice * C_ice * T_initial_ice - m_ice * L_f
Combining like terms:
(T_final) * (m_water * C_water - m_ice * C_ice) = m_water * C_water * T_initial_water - m_ice * C_ice * T_initial_ice - m_ice * L_f
Finally, finding T_final:
T_final = (m_water * C_water * T_initial_water - m_ice * C_ice * T_initial_ice - m_ice * L_f) / (m_water * C_water - m_ice * C_ice)
Let's substitute the values into the equation to find the final temperature of the system:
T_final = (200 g * 4.18 J/g°C * 18°C - 250 g * 2.09 J/g°C * 0°C - 250 g * 334 J/g) / (200 g * 4.18 J/g°C - 250 g * 2.09 J/g°C)
Calculating this expression will give us the final temperature of the system.