Question

What mass of water at \(25.0^{\circ} \mathrm{C}\) added to a Styrofoam cup containing two 50.0 -g ice cubes from a freezer at \(-15.0^{\circ} \mathrm{C}\) will result in a final temperature of \(5.0^{\circ} \mathrm{C}\) for the drink?

Short answer

Answer: 1648 g

Step-by-step solution

Calculate the heat required to raise the temperature of ice cubes to 0°C

We first need to calculate the heat required to raise the temperature of both ice cubes (-15.0°C) to 0°C. We will use the equation \(Q = mcΔT\): \(Q = (m_{ice})(c_{ice})(ΔT_{ice})\) Here, \(m_{ice} = 2 \times 50.0\,\text{g} = 100.0\,\text{g}\) (mass of both ice cubes), \(c_{ice} = 2.093\,\frac{J}{g°C}\) (specific heat capacity of ice), and \(ΔT_{ice} = 15.0\,°\text{C}\) (change in temperature from -15.0°C to 0°C). \(Q_{ice} = (100.0\,\text{g})(2.093\,\frac{J}{g°C})(15.0\,°\text{C}) = 3140\,\text{J}\) (rounded to 3 significant figures)

Calculate the heat required to change ice from solid to liquid

Next, we need to calculate the heat required to change the solid ice cubes into liquid water at 0°C. We will use the equation \(Q = mL\): \(Q = (m_{ice})(L_{fusion})\) Here, \(m_{ice} = 100.0\,\text{g}\) (mass of both ice cubes) and \(L_{fusion} = 334\,\frac{J}{g}\) (latent heat of fusion for water). \(Q_{fusion} = (100.0\,\text{g})(334\,\frac{J}{g}) = 33400\,\text{J}\)

Calculate the heat required to raise the temperature of liquid water to the final temperature

Now, we need to calculate the heat required to raise the temperature of the entire mixture (100 g of liquid water and the mass of water) to the final temperature of 5.0°C. We will use the equation \(Q = mcΔT\): \(Q = (m_{water}+m_{ice})(c_{water})(ΔT_{water})\) Here, \(m_{water} + m_{ice} = 100.0\,\text{g} + m_{water}\) (mass of water added plus the mass of ice cubes), \(c_{water} = 4.18\,\frac{J}{g°C}\) (specific heat capacity of water), and \(ΔT_{water} = 5.0\,°\text{C}\) (change in temperature from 0°C to 5.0°C). \(Q_{water} = (100.0\,\text{g} + m_{water})(4.18\,\frac{J}{g°C})(5.0\,°\text{C}) = (100.0\,\text{g} + m_{water})\times 20.9\,\text{J} \)

Equate the heat lost by the water to the heat gained by the ice

The heat lost by the water (\(Q_{water}\)) will be equal to the heat gained by the ice (\(Q_{total}\)), where \(Q_{total} = Q_{ice} + Q_{fusion}\): \(Q_{water} = Q_{total}\) \((100.0\,\text{g} + m_{water})\times 20.9\,\text{J} = 3140\,\text{J} + 33400\,\text{J}\)

Solve for the mass of water

Now, we can solve for the mass of water (\(m_{water}\)) by simplifying the equation from Step 4: \((100.0\,\text{g} + m_{water})\times 20.9\,\text{J} = 36540\,\text{J}\) Divide both sides by 20.9 J: \(100.0\,\text{g} + m_{water} = 1748\,\text{g}\) (rounded to 3 significant figures) Subtract 100.0 g from both sides: \(m_{water} = 1648\,\text{g}\) Therefore, 1648 g of water at 25.0°C needs to be added to the two 50.0-g ice cubes to achieve a final temperature of 5.0°C for the drink.