Suppose 300 g of water at 20°C is poured over a 50-g cube of ice with a temperature of -6°C. If all the ice melts, what is the final temperature of the water? If all of the ice does not melt, how much ice remains when the water–ice mixture reaches equilibrium?
Look at the first problem below, in "Related Questions"
To find the final temperature of the water when all the ice melts, we can use the principle of conservation of energy. The heat gained by the ice will be equal to the heat lost by the water, using the formula:
\(Q_{\text{ice}} = Q_{\text{water}}\)
The heat gained or lost can be calculated using the specific heat capacity and the equation:
\(Q = mc \Delta T\)
where \(Q\) is the heat gained or lost, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
First, let's calculate the heat gained by the ice. The heat gained by the ice is represented by \(Q_{\text{ice}}\), \(m_{\text{ice}}\) represents the mass of the ice, \(c_{\text{ice}}\) represents the specific heat capacity of ice, and \(\Delta T_{\text{ice}}\) represents the change in temperature of the ice:
\(Q_{\text{ice}} = m_{\text{ice}} c_{\text{ice}} \Delta T_{\text{ice}}\)
Substituting the given values, we have:
\(Q_{\text{ice}} = 50 \, \text{g} \times 2.09 \, \text{J/g°C} \times (0 - (-6))°C\)
Next, let's calculate the heat lost by the water. The heat lost by the water is represented by \(Q_{\text{water}}\), \(m_{\text{water}}\) represents the mass of the water, \(c_{\text{water}}\) represents the specific heat capacity of water, and \(\Delta T_{\text{water}}\) represents the change in temperature of the water:
\(Q_{\text{water}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}}\)
Substitute the given values, we have:
\(Q_{\text{water}} = 300 \, \text{g} \times 4.18 \, \text{J/g°C} \times (\text{final temperature} - 20)°C\)
Since the heat gained by the ice is equal to the heat lost by the water (assuming no heat loss to the surroundings), we can set the two equations equal to each other:
\(Q_{\text{ice}} = Q_{\text{water}}\)
\(50 \, \text{g} \times 2.09 \, \text{J/g°C} \times (0 - (-6))°C = 300 \, \text{g} \times 4.18 \, \text{J/g°C} \times (\text{final temperature} - 20)°C\)
Now we can solve for the final temperature (\(\text{final temperature}\)):
\(12.54 \, \text{J/°C} = 1254 \, \text{J/°C} \times (\text{final temperature} - 20)°C\)
Divide both sides by 1254 J/°C:
\(\text{final temperature} - 20 = \frac{12.54}{1254}\)
\(\text{final temperature} - 20 = 0.01\)
\(\text{final temperature} = 20 + 0.01\)
\(\text{final temperature} = 20.01°C\)
Therefore, the final temperature of the water, when all the ice melts, is 20.01°C.
Now, let's calculate how much ice remains when the water-ice mixture reaches equilibrium. We know that at equilibrium, the final temperature of the system will be the melting point of ice, which is 0°C.
Using the same principle of conservation of energy, we can calculate the heat gained by the ice and the heat lost by the water until they reach equilibrium. We can then use these values to determine how much ice remains.
Following the same initial equations:
\(Q_{\text{ice}} = Q_{\text{water}}\)
\(Q_{\text{ice}} = m_{\text{ice}} c_{\text{ice}} \Delta T_{\text{ice}}\)
\(Q_{\text{water}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}}\)
Substituting the given values and the final temperature as 0°C, we can set up the equation:
\(50 \, \text{g} \times 2.09 \, \text{J/g°C} \times (0 - (-6)) = 300 \, \text{g} \times 4.18 \, \text{J/g°C} \times (0 - 0)\)
Simplifying, we find:
\(62.7 = 0\)
Since this equation is not true, it means that all of the ice will melt, and no ice will remain when the water-ice mixture reaches equilibrium.
Therefore, when the water-ice mixture reaches equilibrium, all the ice will have melted.