Question

If it takes 654 J of energy to warm a 5.51 -g sample of water, how much energy would be required to warm \(55.1 \mathrm{~g}\) of water by the same amount?

Short answer

To warm 55.1 g of water by the same temperature as the 5.51 g sample, it would take approximately 6540 J of energy.

Step-by-step solution

Identify the given information and the target variable.

The given information is: - Energy needed to warm 5.51 g of water (Q1) = 654 J - Mass of the first water sample (m1) = 5.51 g - Mass of the second water sample (m2) = 55.1 g Our goal is to find the energy required to warm the second water sample by the same amount (Q2).

Calculate the specific heat capacity of the water.

First, we need to find the specific heat capacity (c) of water. It is a known value for water and is given by: c = 4.18 J/g°C Now we have the specific heat capacity for water.

Calculate the temperature change for the first water sample.

Next, we will calculate the temperature change (ΔT) for the first water sample using the heat transfer formula: Q = mcΔT Rearrange for ΔT: ΔT = Q/(mc) Now, input the given values and calculate ΔT, using m1 and Q1: ΔT = 654 J/(5.51 g × 4.18 J/g°C) ΔT ≈ 28.47°C The temperature change for the first water sample is 28.47°C.

Calculate the energy needed to warm the second water sample.

Now that we know the temperature change, we can calculate the energy required to warm the second water sample (Q2) using the heat transfer formula and the given mass (m2): Q2 = m2 × c × ΔT Substitute the known values: Q2 = 55.1 g × 4.18 J/g°C × 28.47 °C Q2 ≈ 6540 J So, it would take 6540 J of energy to warm the second water sample (55.1 g) by the same temperature (28.47 °C) as the first water sample.

Key concepts

Specific Heat Capacity

The specific heat capacity of a substance is a crucial concept in thermochemistry. It is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1°C.
This value is unique for each material and provides insight into how different materials absorb heat.

For water, the specific heat capacity is typically 4.18 J/g°C. This relatively high value indicates that water can absorb a lot of heat without a significant change in temperature.

• This property is why water is used as a cooling agent in many systems.
• It's also why large bodies of water, like oceans, have a moderating effect on climate.

Knowing the specific heat capacity of water is vital when performing energy calculations, as it helps determine how much energy is needed for temperature changes during various processes.

Calorimetry

Calorimetry is a branch of thermochemistry that deals with measuring the amount of heat transferred to or from a substance.
It's an empirical method used to determine specific heat capacities, temperature changes, and energy values involved in chemical reactions.

The core idea is to monitor how much heat a sample gains or loses when exposed to energy changes. Using instruments called calorimeters, we can measure these changes quite accurately.

• Calorimeters are insulated to prevent heat exchange with the environment, which ensures accurate measurements.
• The data obtained helps in understanding reaction enthalpies, specific heats, and other thermal properties of substances.

In the exercise, understanding calorimetry helps us figure out how much energy it takes to change the temperature of a water sample. This knowledge facilitates the calculation of energy needs for different masses or conditions.

Energy Calculations

Energy calculations in thermochemistry are essential for predicting how systems respond to heat changes. These calculations figure out how much energy is needed to achieve a desired temperature change in a substance.
The fundamental formula used is:

\[ Q = mc\Delta T \]
Here, \(Q\) is the heat energy transferred, \(m\) is the mass of the substance, \(c\) is its specific heat capacity, and \(\Delta T\) is the temperature change. Using this formula, we can predict or measure the energy requirements for heating or cooling substances.

• In the step-by-step solution, the formula helps calculate how much energy is required to warm different masses of water under identical conditions.
• This method is widely applied in many fields, including chemistry, physics, and engineering, to design and optimize thermal systems.

For the exercise in question, understanding energy calculations allows us to scale up the energy requirement from a smaller sample of water to a larger one while keeping the temperature change constant.