An ice cube with a mass of \(25 \mathrm{~g}\) at \(-18{ }^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds $250 \mathrm{~mL}\( of hot water, initially at \)85^{\circ} \mathrm{C}$. What is the final temperature in the cup? The density of liquid water is $1.00 \mathrm{~g} / \mathrm{mL}\(; the specific heat capacity of ice is \)2.03 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C} ;$ the specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}-\mathrm{K} ;\) the enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).

To calculate the energy required to heat the ice from -18°C to 0°C, we use the formula for energy transfer: \(q = mc\Delta T\) Where: \(q\) = energy transfer (J), \(m\) = mass of the ice (g), \(c\) = specific heat capacity of the ice (J/g°C), and \(\Delta T\) = change in temperature (°C). Here, \(m = 25 g\), \(c = 2.03 J/g°C\), and \(\Delta T = 0°C - (-18°C) = 18 °C\). Thus, the energy required to heat the ice to 0°C is: \(q_1 = (25 g)(2.03 J/g °C)(18 °C) = 913.5 J\)

To calculate the energy required to melt the ice at 0°C, we use the enthalpy of fusion (\(Qf\)). Given that the enthalpy of fusion of water is \(6.01 kJ/mol\), we first need to determine the number of moles in the ice cube: Number of moles = \(\frac{mass}{molar\,mass}\) Since the molar mass of water is 18 g/mol, the number of moles in 25g of ice is: \(\frac{25 g}{18 g/mol} = 1.389 mol\) Now, we can find the energy required to melt the ice: \(q_2 = nQ_f = (1.389 mol)(6.01 kJ/mol) = 8.35 kJ = 8350 J\)

To solve for the final temperature, we equate the energy absorbed in heating the ice cube to the energy released by the hot water. The energy absorbed is the sum of energies (\(q_1\) and \(q_2\)) from steps 1 and 2. Let T be the final temperature, then the energy released by hot water is given by the formula: \(q_3 = m_{water}c_{water}\Delta T_{water}\) The mass of hot water can be determined from its volume and density: \(m_{water} = 250 \,mL * \frac{1.00\,g}{1\,mL} = 250\,g\) And, \(\Delta T_{water} = T_{initial} - T_{final} = 85 °C - T\) Hence, the energy loss by the hot water can be represented as: \(q_3 = (250 g)(4.184 J/g °C)(85 °C - T)\) Now we equate the energy absorbed in heating the ice cube (sum of \(q_1\) and \(q_2\)) to the energy released by the hot water (\(q_3\)): \(q_1 + q_2 = q_3\) \(913.5 J + 8350 J = (250 g)(4.184 J/g °C)(85 °C - T)\) Now, solving the equation for the final temperature, T: \((85 °C - T) = \frac{913.5 J + 8350 J}{(250 g)(4.184 J/g °C)}\) \((85 °C - T) ≈ 9.818 °C\) Upon finding the difference: \(T ≈ 85 °C - 9.818 °C ≈ 75.18 °C\) So, the final temperature in the cup is approximately \(75.18 °C\).