Question

A refrigerator with a coefficient of performance of 3.80 is used to \(\operatorname{cool} 2.00 \mathrm{~L}\) of mineral water from room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) to \(4.00^{\circ} \mathrm{C} .\) If the refrigerator uses \(480 . \mathrm{W}\) how long will it take the water to reach \(4.00^{\circ} \mathrm{C}\) ? Recall that the heat capacity of water is \(4.19 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K}),\) and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\). Assume the other contents of the refrig. erator are already at \(4.00^{\circ} \mathrm{C}\).

Short answer

Answer: Approximately 96.7 seconds.

Step-by-step solution

Calculate mass of water

First, we need to find the mass of water, and given the density of water, we can calculate the mass, using the formula mass = volume × density. Given 2L or 2000 cm³ water and density of water is 1g/cm³. Mass of water (m) = Volume × Density = 2000 cm³ × 1 g/cm³ = 2000 g = 2 kg

Calculate the heat required to change the water temperature

Now we need to calculate the heat required to change the temperature of water from 25°C to 4°C. We will use the formula for heat transfer: Q = m × c × ΔT, where Q is the heat transfer, m is the mass of water in kg, c is the specific heat capacity of water, and ΔT is the temperature change. Given heat capacity c = 4.19 kJ/(kg K), initial temperature T1 = 25°C, and final temperature T2 = 4°C. ΔT = T2 - T1 = 4°C - 25°C = -21°C Q = m × c × ΔT = 2 kg × 4.19 kJ/(kg K) × (-21°C) = -176.38 kJ Since cooling, all values are negative. So, the heat required(Q) = 176.38 kJ.

Define the coefficient of performance of the refrigerator

The coefficient of performance (COP) of the refrigerator is given by the formula COP = Q_cooled / W where Q_cooled is the amount of heat extracted from the water in kJ, and W is the work done in kJ. Given COP = 3.80.

Calculate the work required to extract the heat from the water

We can rearrange the formula for COP and solve for W: W = Q_cooled / COP = 176.38 kJ / 3.80 = 46.415 kJ

Calculate the time required

Now we are given that the refrigerator uses 480W of power to do 46.415kJ work. We can calculate the time (t) required using the formula: Power(P) = work(W) / time(t) Power is given in Watts, which is equivalent to J/s. So first, we need to convert the 46.415 kJ to J (Joules): 46.415 kJ = 46,415 J Now, given the power (P) = 480 W, let's calculate the time (t): t = W / P = 46,415 J / 480 J/s ≈ 96.7 seconds So, it takes approximately 96.7 seconds for the refrigerator to cool the water from 25.0°C to 4.00°C.

Key concepts

Coefficient of Performance

When it comes to thermodynamics in refrigeration, the Coefficient of Performance (COP) is a key concept. It is a measure of a refrigerator's efficiency and indicates how well the unit can move heat versus the amount of work or energy it uses to do so. Specifically, COP is the ratio of the heat removed from the refrigerated space to the work input required to remove it. A higher COP means greater efficiency, as the refrigerator can remove more heat with less energy.

In the given problem, a refrigerator with a COP of 3.80 implies that for every unit of energy used by the refrigerator, it can remove 3.80 units of heat from the water. Understanding COP helps us determine how long it will take for the refrigerator to cool the mineral water based on its power consumption, which is essential for both manufacturers designing appliances and consumers looking to reduce their energy usage.

Heat Capacity of Water

The Heat Capacity of Water plays a significant role in calculating the energy needed to change water's temperature. Heat capacity, often designated as 'c', is defined as the amount of heat energy required to raise the temperature of a given mass of a substance by one degree Celsius. The heat capacity of water is high compared to many other substances, meaning water can absorb a lot of heat before its temperature rises significantly - an important characteristic that makes water an excellent coolant.

For water, the heat capacity is 4.19 kJ/kg K, which signifies the amount of energy needed to raise the temperature of one kilogram of water by one Kelvin (or one degree Celsius). This property is crucial when we calculate the time it will take to cool a certain volume of water, as seen in our exercise. The heat capacity is a cornerstone concept in thermodynamics, as it determines how a substance will respond to the addition or removal of heat.

Energy Transfer

In our exercise, we observe an Energy Transfer from the water to the refrigeration system. Energy transfer is a core principle of thermodynamics reflecting the movement of energy from one place to another. In refrigeration, energy in the form of heat is transferred from the interior of the refrigerator (and the items within, like our water) to the exterior environment. This is how a refrigerator keeps things cool - by removing heat energy from inside.

In the calculation provided, the heat energy (Q) is calculated as the product of the mass of the water, its specific heat capacity, and the change in temperature (ΔT). The transfer of this energy corresponds to the cooling of the water – from a 'higher-energy' warm state to a 'lower-energy' cool state. Understanding this process is essential for grasifying the physical changes occurring within the refrigerator and determining the time needed for the refrigeration cycle.

Specific Heat Capacity

Closely related to the general concept of heat capacity is Specific Heat Capacity. This property is the amount of heat needed to raise the temperature of one gram (or one kilogram) of a substance by one degree Celsius (or one Kelvin). Specific heat capacity is a fundamental concept when discussing heat transfer in thermodynamic processes because it describes how different materials react to heat.

For water, the specific heat capacity is quite high, which historically has made it useful for heating and cooling applications. In the context of our exercise, we use the specific heat capacity of water to determine the quantity of heat that must be removed to cool our 2.00 L of mineral water from 25.0°C to 4.00°C. By understanding the specific heat capacity, we can predict how much energy is required to achieve a certain temperature change in a substance, which in this case was crucial for finding out how long the refrigeration process would take.