(a) What amount of heat (in joules) is required to raise the temperature of $1 \mathrm{~g}$ of water by 1 kelvin? (b) What amount of heat (in joules) is required to raise the temperature of 1 mole of water by 1 kelvin? (c) What is the heat capacity of \(370 \mathrm{~g}\) of liquid water? (d) How many kJ of heat are needed to raise the temperature of $5.00 \mathrm{~kg}\( of liquid water from 24.6 to \)46.2^{\circ} \mathrm{C} ?$

Using the heat equation given above: \[Q = mc\Delta T\] The mass of water, \(m\) is 1 g, the specific heat capacity of water, \(c\) is \(4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\) and the change in temperature, \(\Delta T\) is 1 K. Substituting these values into the equation, we get: \[Q = 1 \times 4.18 \times 1\] \[Q = 4.18~ \mathrm{J}\] So, the amount of heat required is 4.18 J.

First, we need to find the mass of 1 mole of water. The molecular weight of water is approximately 18 g/mol. Therefore, 1 mole of water weighs 18 g. Now, using the heat equation: \[Q = mc\Delta T\] The mass of water, \(m\) is 18 g, the specific heat capacity of water, \(c\) is \(4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\) and the change in temperature, \(\Delta T\) is 1 K. Substituting these values into the equation, we get: \[Q = 18 \times 4.18 \times 1\] \[Q = 75.24~ \mathrm{J}\] So, the amount of heat required is 75.24 J.

The heat capacity of a substance is given by the product of its mass and its specific heat capacity: \[C = mc\] Where \(C\) is the heat capacity, \(m\) is the mass, and \(c\) is the specific heat capacity. For 370 g of water, the specific heat capacity, \(c\) is \(4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\). So, the heat capacity of 370 g of water is: \[C = 370 \times 4.18\] \[C = 1546.6~\frac{\mathrm{J}}{\mathrm{K}}\] So, the heat capacity of 370 g of liquid water is 1546.6 J/K.

First, we need to find the change in temperature, \(\Delta T\). The initial temperature is 24.6°C, and the final temperature is 46.2°C. Therefore, the change in temperature is: \[\Delta T = T_{f} - T_{i} = 46.2 - 24.6 = 21.6~\mathrm{K}\] Next, we use the heat equation: \[Q = mc\Delta T\] The mass of water, \(m\) is 5.00 kg, the specific heat capacity of water, \(c\) is \(4186 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\) and the change in temperature, \(\Delta T\) is 21.6 K. Substituting these values into the equation, we get: \[Q = 5.00 \times 4186 \times 21.6\] \[Q = 453124.8 ~\mathrm{J}\] To convert this to kJ, divide by 1000: \[Q = \frac{453124.8}{1000}\] \[Q = 453.1~ \mathrm{kJ}\] So, 453.1 kJ of heat is needed to raise the temperature of 5.00 kg of liquid water from 24.6°C to 46.2°C.