Question

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

Short answer

(a) 4.18 J (b) 75.24 J (c) 1546.6 J/K (d) 453.1 kJ

Step-by-step solution

(a) Calculate the heat required to raise the temperature of 1g of water by 1K

Using the heat equation given above: $$Q=mc\mathrm{\Delta }T$$ 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, $\mathrm{\Delta }T$ 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.

(b) Calculate the heat required to raise the temperature of 1 mole of water by 1K

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: $$Q=mc\mathrm{\Delta }T$$ 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, $\mathrm{\Delta }T$ 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.

(c) Calculate the heat capacity of 370g of liquid water

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.

(d) Calculate the heat required to raise the temperature of 5.00 kg of liquid water from 24.6°C to 46.2°C

First, we need to find the change in temperature, $\mathrm{\Delta }T$. The initial temperature is 24.6°C, and the final temperature is 46.2°C. Therefore, the change in temperature is: $$\mathrm{\Delta }T=T_f-T_i=46.2-24.6=21.6\mathrm{K}$$ Next, we use the heat equation: $$Q=mc\mathrm{\Delta }T$$ 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, $\mathrm{\Delta }T$ 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.

Key concepts

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat energy is required to raise the temperature of a unit mass of a substance by one unit of temperature. For water, this value is relatively high at 4.18 J/g·K, making it exemplary as a heat storage medium. This is why water is extensively used in heating and cooling systems.

Understanding specific heat capacity can help explain many natural and engineered thermal processes. Some key points are:

• Different substances have different specific heat capacities, meaning they will heat up and cool down at different rates.
• Materials with high specific heat capacities tend to retain heat longer than those with lower values.
• In calculations involving temperature change, the specific heat capacity allows for precise determination of the energy needed for a given temperature change.

Next time you heat a pot of water, consider how much energy is being absorbed by the water due to its specific heat capacity.

Molecular Weight

The molecular weight of a substance is the mass of one mole of its molecules. It provides a bridge between microscopic and macroscopic scales. For a substance like water (H₂O), the molecular weight is approximately 18 g/mol. This means one mole of water has a mass of 18 grams.

Molecular weight is crucial in stoichiometry and various chemical calculations, especially when scaling small quantities to more tangible, practical amounts.

• Molecular weight serves as a conversion factor between the number of molecules and the mass of a sample.
• In many experiments or reactions, knowing the molecular weight allows chemists to measure exact mole quantities.
• It's commonly used with the ideal gas law and in determining molar concentrations in solutions.

For water, the relationship between mass and molecular weight allows us to calculate heat requirements or other properties for any given amount of water molecules.

Heat Equation

The heat equation $$Q=mc\mathrm{\Delta }T$$is fundamental in thermodynamics for quantifying the amount of thermal energy transferred to or from a substance. In this equation:

• $Q$ stands for the heat energy transferred.
• $m$ is the mass of the substance.
• $c$ represents the specific heat capacity.
• $\mathrm{\Delta }T$ is the temperature change, calculated as the final temperature minus the initial temperature.

This formula is incredibly versatile and widely used across various scientific disciplines to solve problems involving heat transfer.

For students, mastering this equation is critical for doing well in physics and chemistry courses that deal with thermodynamic systems. Whether you're looking to determine how much energy is needed to heat your morning coffee or to design a heat engine, the heat equation is a powerful tool that connects these everyday experiences to the laws of physics. Always remember, it all comes back to knowing the mass, specific heat capacity, and temperature change!