The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) is $2.22 \mathrm{~J} / \mathrm{g}\( -K. \)(\mathbf{a})$ How many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from 10.0 to \(25.0^{\circ} \mathrm{C} ?(\mathbf{b})\) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of $\mathrm{C}_{8} \mathrm{H}_{18}(l)\( by a certain amount or increasing the temperature of \)1 \mathrm{~mol}\( of \)\mathrm{H}_{2} \mathrm{O}(l)$ by the same amount?

We are given the mass of octane (\(m = 80.0\) g), specific heat of octane (\(c = 2.22\) J/g-K), and the initial (\(T_1 = 10.0°C\)) and final (\(T_2 = 25.0°C\)) temperatures. We can now calculate the heat required using the formula: \(q = mcΔT\). 1. Calculate the change in temperature: \(ΔT = T_2 - T_1 = 25.0°C - 10.0°C = 15.0°C\). 2. Calculate the heat required: \(q = (80.0\text{ g})(2.22\text{ J/g-K})(15.0\text{ K}) = 2664\text{ J}\). So, the heat required to raise the temperature of 80.0 g of octane from 10.0 to 25.0°C is 2664 J.

First, we need to find the molar mass of octane \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\): 1. Molar mass of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\): \((8 \times 12.01\text{ g/mol}) + (18 \times 1.01\text{ g/mol}) = 114.22\text{ g/mol}\) Now, let's assume the change in temperature for both octane and water is \(ΔT\): 2. Heat required for 1 mole of octane (\(q_1\)): \(q_1 = (1\text{ mol})(114.22\text{ g/mol})(2.22\text{ J/g-K})ΔT = 253.47ΔT\text{ J}\) 3. Heat required for 1 mole of water (\(q_2\)): \(q_2 = (1\text{ mol})(18.02\text{ g/mol})(4.18\text{ J/g-K})ΔT = 75.33ΔT\text{ J}\) Comparing the heat required for 1 mole of octane and 1 mole of water: 4. Check if \(q_1 > q_2\): \((253.47ΔT\text{ J}) > (75.33ΔT\text{ J})\) Since the heat required for 1 mole of octane (\(q_1\)) is greater than the heat required for 1 mole of water (\(q_2\)), increasing the temperature of 1 mole of octane by the same amount requires more heat than increasing the temperature of 1 mole of water.