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A system consisting of 82.5 g of liquid water at \(300 .\) K is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of 1.75 A passes through the 25.0 ohm resistor for \(100 .\) s, what is the final temperature of the water?

Short Answer

Expert verified
The final temperature of the water can be found by following these steps: 1. Calculate the electrical power (P) provided to the water using \(P = I^2 \times R\), where I = 1.75 A and R = 25.0 Ω. 2. Calculate the energy (E) provided to the water using \(E = P \times t\), where t = 100 s. 3. Calculate the change in temperature (ΔT) using \(\Delta T = \frac{E}{m \times c}\), where m = 82.5 g and c = 4.18 J/(g·K). 4. Find the final temperature (T2) by adding ΔT to the initial temperature (T1 = 300 K). By following these steps and the given values, the final temperature of the water can be calculated.

Step by step solution

01

Calculate the electrical power

First, we need to calculate the electrical power provided to the water by the immersion heater. We can use the formula: \(P = I^2 \times R\) Where: P = power in watts (W), I = current in amperes (A), R = resistance in ohms (Ω). Plug in the given values, I = 1.75 A and R = 25.0 Ω \(P = (1.75)^2 \times 25.0\)
02

Calculate the energy provided

Next, we have to find the total amount of energy provided to the water during the heating time using the formula: \(E = P \times t\) Where: E = energy in joules (J), P = power in watts (W), t = time in seconds (s). Plug in the calculated power from Step 1 and the given time (t = 100 s) to calculate the energy provided to the water. \(E = P \times 100\)
03

Calculate the energy required to change the water temperature

The energy required to change the temperature of the water can be calculated using the formula: \(E = m \times c \times \Delta T\) Where: E = energy in joules (J), m = mass in grams (g), c = specific heat capacity of water, c = 4.18 J/(g·K), ΔT = change in temperature. Rearranging the formula for ΔT, we get: \(\Delta T = \frac{E}{m \times c}\) Now, we can plug in the energy calculated in step 2, the mass (m = 82.5 g), and the specific heat capacity of water (c = 4.18 J/(g·K)).
04

Calculate the final temperature

Finally, to calculate the final temperature, we need to add the change in temperature (ΔT) to the initial temperature (T1 = 300 K). \(T_2 = T_1 + \Delta T\) Plug in the initial temperature (T1 = 300 K) and the calculated change in temperature (ΔT). Now, we have all the steps required to find the final temperature (T2) of the water after heating by the immersion heater. Following the steps and calculations mentioned, we can calculate the final temperature of the water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
The concept of specific heat capacity is fundamental in understanding how the temperature of a substance changes when it absorbs or releases heat. It is a measure of how much heat energy (in joules) is needed to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin).

Water has a relatively high specific heat capacity of 4.18 J/(g·K), which means it requires a significant amount of energy to change its temperature. This is why we see water used in various applications for heating and cooling—it has the ability to absorb and release large quantities of energy without undergoing extreme temperature changes. When solving for temperature changes in water, as shown in the exercise, we use this specific value to determine how the water's temperature will be affected by the energy transferred from the immersion heater.

Understanding the specific heat capacity enables us to calculate the energy required to heat or cool substances to a desired temperature, important for both theoretical exercises and practical applications in the real world.
Energy Transfer
Energy transfer is the process of energy moving from one place or object to another. In our exercise, energy is transferred in the form of heat from the electric immersion heater to the water. The transfer occurs because of the electrical resistance of the immersion heater, which converts electrical energy into heat energy through the resistor.

Once the electrical power is known, as calculated in the solution's Step 1, one can determine the total amount of energy transferred over a specific time by multiplying the power by the time. This is effectively how we calculate the energy that is then used by the water to increase its temperature (Step 2 of the solution).

The ability to predict and calculate energy transfer enables us to control the temperature changes in a system, which is crucial in fields such as chemical engineering, environmental science, and many industrial processes.
Electrical Power
Electrical power, measured in watts (W), refers to the rate at which electrical energy is converted into another form of energy, such as heat, light, or motion. The formula to calculate the power consumed by an electrical device is \( P = I^2 \times R \), where \( I \) is the current in amperes, and \( R \) is the resistance in ohms.

During the heating process in our textbook problem, the immersion heater's resistance causes the electrical energy to be converted into heat energy at a rate calculated using this formula. The higher the resistance or the current, the more power is consumed, and consequently, more heat is produced (Step 1). By knowing the electrical power, one can manage the amount of energy being fed into a system, as seen in Step 2 of the solution, which is vital for designing and operating electrical heating systems effectively.
Heat Transfer in Physical Chemistry
Heat transfer is a key concept in physical chemistry that involves the movement of thermal energy from one physical system to another. It is central to processes like boiling, melting, or chemical reactions. In the context of our exercise, we're interested in how heat transfer affects the temperature of a system, specifically water being heated by an immersion heater.

The specific heat equation \( E = m \times c \times \Delta T \) outlines the relationship between the energy added to the system (\( E \) in joules) and the resultant temperature change (\( \Delta T \) in Kelvin), with \( m \) representing the mass of the substance and \( c \) the specific heat capacity. By manipulating this equation, you can solve for any of the unknown variables given the others (as demonstrated in Step 3 of the solution).

The concept of heat transfer is crucial not only for solving textbook problems but also for the analysis and design of industrial processes, the management of thermal systems, and offering insights into the thermal behavior of chemical reactions.

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Most popular questions from this chapter

The heat capacity of solid lead oxide is given by \\[ C_{P, m}=44.35+1.47 \times 10^{-3} \frac{T}{\mathrm{K}} \text { in units of } \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1} \\] Calculate the change in enthalpy of 1.75 mol of \(\mathrm{PbO}(s)\) if it is cooled from 825 K to 375 K at constant pressure.

At \(298 \mathrm{K}\) and 1 bar pressure, the density of water is \(0.9970 \mathrm{g} \mathrm{cm}^{-3},\) and \(C_{P, m}=75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The change in volume with temperature is given by \(\Delta V=V_{\text {initial}} \beta \Delta T\) where \(\beta,\) the coefficient of thermal expansion, is \(2.07 \times 10^{-4} \mathrm{K}^{-1}\). If the temperature of \(325 \mathrm{g}\) of water is increased by \(25.5 \mathrm{K}\), calculate \(w, q, \Delta H,\) and \(\Delta U\)

A 1.75 mole sample of an ideal gas for which \(C_{V, m}=20.8 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) is heated from an initial temperature of \(21.2^{\circ} \mathrm{C}\) to a final temperature of \(380 .^{\circ} \mathrm{C}\) at constant volume. Calculate \(q, w, \Delta U,\) and \(\Delta H\) for this process.

Calculate \(\Delta H\) and \(\Delta U\) for the transformation of \(2.50 \mathrm{mol}\) of an ideal gas from \(19.0^{\circ} \mathrm{C}\) and 1.00 atm to \(550 .^{\circ} \mathrm{C}\) and \(19.5 \mathrm{atm}\) if \(C_{P, m}=20.9+0.042 \frac{T}{\mathrm{K}}\) in units of \(\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}\)

One mole of an ideal gas is subjected to the following changes. Calculate the change in temperature for each case if \(C_{V, m}=3 R / 2\) a. \(q=-425 \mathrm{J}, w=185 \mathrm{J}\) b. \(q=315 .\) J, \(w=-315\) \(\mathbf{c} . q=0, w=225 \mathrm{J}\)

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