(a) What is the specific heat of liquid water? (b) What is the molar heat capacity of liquid water? (c) What is the heat capacity of \(185 \mathrm{~g}\) of liquid water? (d) How many \(\mathrm{kJ}\) of heat are needed to raise the temperature of \(10.00 \mathrm{~kg}\) of liquid water from \(24.6^{\circ} \mathrm{C}\) to \(46.2^{\circ} \mathrm{C} ?\)

The specific heat of liquid water is \(4.18 \,\text{J/g}^\circ\text{C}\). This value is experimentally determined and can be found in reference materials such as textbooks and online resources.

To find the molar heat capacity of liquid water, we need to multiply the specific heat by the molar mass of water. The molar mass of water is approximately 18.015 g/mol. Molar heat capacity = Specific heat × Molar mass Molar heat capacity = \(4.18 \,\text{J/g}^\circ\text{C} \times 18.015\,\text{g/mol}\) Molar heat capacity ≈ \(75.33 \,\text{J/mol}^\circ\text{C}\)

To calculate the heat capacity of 185 g of liquid water, we use the formula: Heat capacity = Mass × Specific heat Heat capacity = \(185\,\text{g} \times 4.18 \,\text{J/g}^\circ\text{C}\) Heat capacity ≈ \(773.3 \,\text{J/}^\circ\text{C}\)

To calculate the heat needed to raise the temperature of 10 kg of liquid water from 24.6°C to 46.2°C, we use the heat transfer formula: Q = mcΔT where Q is the heat needed, m is the mass of water, c is the specific heat, and ΔT is the change in temperature. First, we need to convert the mass of water from kg to g: Mass of water = \(10,000\,\text{g}\) (since 1 kg = 1000 g) Next, we calculate the change in temperature: ΔT = T_final - T_initial ΔT = 46.2°C - 24.6°C ΔT = 21.6°C Now, we can plug in the values into the heat transfer formula: Q = \(10,000\,\text{g} \times 4.18 \,\text{J/g}^\circ\text{C} \times 21.6^\circ\text{C}\) Q ≈ \(903,360 \,\text{J}\) To convert the heat needed to kilojoules, we divide by 1000: Q ≈ \(903.36 \,\text{kJ}\)