A 13.0-g sample of ice at -13.0°C is mixed with 105.0 g of water at 78.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g · °C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
q to raise temperature of ice from -13C to zero C (but remain as ice).
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q to melt ice at zero C and remain at zero C.
q2 = mass ice x heat fusion
q to raise melted ice from zero C to final T.
q3 = mass melted ice x specific heat water x (Tfinal-Tinitial)
q to lower temperature of hot water to final T.
q4 = mass hot water x specific heat x (Tfinal-Tinitial).
q1 + q2 + q3 + q4 = 0
Substitute each portion into q1 + q2 .... and solve for Tfinal.
To calculate the final temperature of the mixture, we need to consider the heat gained or lost by each component.
First, let's calculate the heat gained by the ice as it warms up to the final temperature (Tf). We can use the formula:
q_ice = m_ice * C_ice * (Tf - Ti)
where:
q_ice = heat gained by the ice (in joules)
m_ice = mass of ice (13.0 g)
C_ice = heat capacity of ice (2.08 J/g · °C)
Ti = initial temperature of ice (-13.0°C)
q_ice = 13.0 g * 2.08 J/g · °C * (Tf - (-13.0°C))
q_ice = 13.0 g * 2.08 J/g · °C * (Tf + 13.0°C)
Next, let's calculate the heat lost by the water as it cools down to the final temperature (Tf). We can use the formula:
q_water = m_water * C_water * (Tf - Ti)
where:
q_water = heat lost by the water (in joules)
m_water = mass of water (105.0 g)
C_water = heat capacity of water (4.18 J/g · °C)
Ti = initial temperature of water (78.0°C)
q_water = 105.0 g * 4.18 J/g · °C * (Tf - 78.0°C)
Now, since we assume no heat loss to the surroundings, the heat gained by the ice must be equal to the heat lost by the water. So we can set up the following equation:
q_ice = q_water
13.0 g * 2.08 J/g · °C * (Tf + 13.0°C) = 105.0 g * 4.18 J/g · °C * (Tf - 78.0°C)
Now, let's solve for Tf by isolating it on one side of the equation.
13.0 g * 2.08 J/g · °C * Tf + 13.0 g * 2.08 J/g · °C * 13.0°C = 105.0 g * 4.18 J/g · °C * Tf - 105.0 g * 4.18 J/g · °C * 78.0°C
(13.0 g * 2.08 J/g · °C - 105.0 g * 4.18 J/g · °C) * Tf = 105.0 g * 4.18 J/g · °C * 78.0°C - 13.0 g * 2.08 J/g · °C * 13.0°C
Tf = (105.0 g * 4.18 J/g · °C * 78.0°C - 13.0 g * 2.08 J/g · °C * 13.0°C) / (13.0 g * 2.08 J/g · °C - 105.0 g * 4.18 J/g · °C)
Simplifying the expression further, we get:
Tf ≈ 52.21°C
Therefore, the final temperature of the mixture is approximately 52.21°C.