Problem 49 - Chemistry: The Central Science

Question

(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 \(46.2^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

(a) The specific heat of liquid water is \(c_{water} = 4.18 \frac{J}{g^\circ C}\). (b) The molar heat capacity of liquid water is \(c_{molar} = 75.33 \frac{J}{mol^\circ C}\). (c) The heat capacity of 185g of liquid water is \(Q = 773.3\, J/^\circ C\). (d) The heat needed to raise the temperature of 10.00 kg of liquid water from 24.6°C to 46.2°C is 902.04 kJ.

Step 1: (a) Specific heat of liquid water

The specific heat of liquid water is a known property, which is: \(c_{water} = 4.18 \frac{J}{g^\circ C}\)

Step 2: (b) Molar heat capacity of liquid water

To find the molar heat capacity, we first need to know the molar mass of water. The molecular formula of water is H2O, with molar mass: \(H = 1.01 \, g/mol\) \(O = 16.00 \, g/mol\) So the molar mass of water is: M = 2(1.01) + 16.00 = 18.02 g/mol Now we can find the molar heat capacity by multiplying the specific heat by the molar mass: \(c_{molar} = c_{water} \cdot M = 4.18 \frac{J}{g^\circ C} \cdot 18.02 \, g/mol = 75.33 \frac{J}{mol^\circ C}\)

Step 3: (c) Heat capacity of 185g liquid water

To find the heat capacity of 185g of liquid water, we will use the specific heat formula: \(Q = mc\) Where Q is the heat capacity, m is the mass of the substance (185g), and c is the specific heat (4.18 J/g°C). Multiplying the mass and specific heat, we get: \(Q = (185\,g) \times (4.18 \frac{J}{g^\circ C}) = 773.3\, J/^\circ C\)

Step 4: (d) Heat needed to raise temperature from 24.6°C to 46.2°C

To find the heat needed to raise the temperature of 10.00 kg of liquid water from 24.6°C to 46.2°C, we will use the formula for heat capacity: \(Q = mcΔT\) where ΔT is the change in temperature (46.2°C - 24.6°C), m is the mass (10.00 kg = 10000 g), and c is the specific heat (4.18 J/g°C). Plugging in the values, we get: \(Q = (10000\,g) \times (4.18 \frac{J}{g^\circ C}) \times (46.2^\circ C - 24.6^\circ C) = 902040 \, J\) Now we convert J to kJ: \(902040\,J = 902.04 \,kJ\) So, the heat needed is 902.04 kJ.

Key Concepts

Molar Heat Capacity of Water

When studying the heating or cooling of substances, particularly water, we come across the concept of molar heat capacity. This is the amount of heat energy required to raise the temperature of one mole of a substance by one degree Celsius. It's an essential concept in chemistry because it accounts for the amount of substance present when considering heat transfer.

For water, the molar heat capacity is particularly significant because water's high capacity indicates its ability to moderate temperature changes in its environment, a characteristic that contributes significantly to Earth's climate stability. To calculate this, we use the molar mass of water, which is approximately 18.02 grams per mole, combining hydrogen's contribution (2 x 1.01 g/mol) and oxygen's (16.00 g/mol). Multiplying this molar mass by water's specific heat (\(4.18 \frac{J}{g^\text{o}C}\)), yields a molar heat capacity of approximately \(75.33 \frac{J}{mol^\text{o}C}\), a value intrinsic to all water molecules regardless of the amount of substance. Such calculations are not only crucial for academic purposes but also for real-world applications such as designing thermal systems and understanding environmental phenomena.

Heat Capacity Calculation

To really grasp the concept of heat capacity, let’s dive into the practical side of it. Heat capacity is the quantity of heat (usually in joules or calories) needed to change the temperature of a certain amount of a substance by one degree Celsius. In a typical classroom problem, if you're given a mass of water, say 185 grams, and its specific heat (\(4.18 \frac{J}{g^\text{o}C}\)), you can seamlessly find its heat capacity.

You simply multiply the mass of the substance by its specific heat. In this case, \(185 \text{g} \times 4.18 \frac{J}{g^\text{o}C} = 773.3 J/^\text{o}C\), which is the heat capacity. This means it will take 773.3 joules of energy to raise the temperature of 185 grams of water by one degree Celsius. This is particularly useful to know in heating and cooling systems, where the efficiency and energy requirements are paramount to design and operation.

Thermal Energy Transfer

Thermal energy transfer, or the movement of heat from one place to another, is a fundamental property of thermodynamics. It can take place through processes such as conduction, convection, and radiation. In the context of heating water, we usually deal with conduction, where heat energy is directly transferred through the material.

To quantify this transfer, we use the formula \(Q = mc\triangle{T}\), where 'Q' represents the thermal energy transferred, 'm' is the mass of the substance, 'c' is the specific heat, and \triangle{T} is the change in temperature. Here’s a practical example: If you need to heat 10 kg of liquid water from 24.6°C to 46.2°C, you calculate the energy transfer by first converting the mass to grams for consistency with the specific heat unit. With 10 kg equivalent to 10000 grams and the specific heat at \(4.18 \frac{J}{g^\text{o}C}\), the change in temperature (ΔT) being 21.6°C, the total energy required is found to be 902.04 kJ. This is critical for understanding heat transfer efficiency and is extensively applied in fields such as engineering, environmental science, and energy management.