Urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), is used in the manufacture of resins and glues. When \(5.00 \mathrm{~g}\) of urea is dissolved in \(250.0 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\) at \(30.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, \(27.6 \mathrm{~kJ}\) of heat is absorbed. (a) Is the solution process exothermic? (b) What is \(q_{\mathrm{H}_{2} \mathrm{O}}\) ? (c) What is the final temperature of the solution? (Specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (d) What are the initial and final temperatures in \({ }^{\circ} \mathrm{F}\) ?

Since the heat is absorbed (instead of being released) during the solution process, the process is endothermic (not exothermic).

Since the process is endothermic and absorbs 27.6 kJ (or 27600 J) of heat, the heat absorbed by the water during the solution process can be represented as: \(q_{H_2O} = 27600 \ \text{J}\).

To find the final temperature of the solution, we can use the formula for heat transfer: \(q = mcΔT\), where \(q\) is the heat absorbed or released, \(m\) is the mass of the substance, \(c\) is the specific heat, and \(ΔT\) is the change in temperature. In this case, the mass of water (\(m\)) is 250.0 g (since the density of water is 1.00 g/mL), the initial temperature (\(T_i\)) is 30.0°C, and the specific heat of water (\(c\)) is 4.18 J/g·°C. We can rearrange the formula to solve for the change in temperature (\(ΔT\)): \(ΔT = \frac{q}{mc} = \frac{27600 \ \text{J}}{250.0 \ \text{g} \cdot 4.18 \ \text{J/g} \cdot{ }^{\circ} \mathrm{C}} = \frac{27600}{1045} = 26.4{ }^{\circ} \mathrm{C}\) Now we can find the final temperature (\(T_f\)) by adding the change in temperature to the initial temperature: \(T_f = T_i + ΔT = 30.0{ }^{\circ} \mathrm{C} + 26.4{ }^{\circ} \mathrm{C} = 56.4{ }^{\circ} \mathrm{C}\)

To convert the initial and final temperatures from °C to °F, we can use the formula: \(T_{(°F)} = T_{(°C)} \cdot \frac{9}{5} + 32\) For the initial temperature: \(T_i{(}^\circ F{)} = 30.0{ }^\circ C \cdot \frac{9}{5} + 32 = 54{ }^\circ F\) For the final temperature: \(T_f{(}^{\circ}F{)} = 56.4{ }^{\circ} \mathrm{C} \cdot \frac{9}{5} + 32 = 133.5{ }^{\circ} \mathrm{F}\)