A Suppose that only two 45-g ice cubes had been added to...

A Suppose that only two $45-g$ ice cubes had been added to your glass containing $5.00 \times 10^{2} \mathrm{mL}$ of tea (see Study Question 78). When thermal equilibrium is reached, all of the ice will have melted, and the temperature of the mixture will be somewhere between $20.0^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C} .$ Calculate the final temperature of the beverage. (Note: The $90 \mathrm{g}$ of water formed when the ice melts must be warmed from $0^{\circ} \mathrm{C}$ to the final temperature.)

Key Concepts

Latent Heat of Fusion

Latent heat of fusion is the amount of heat required to convert a unit mass of a substance from solid to liquid at its melting point without a change in temperature. This property is essential when dealing with phase changes, such as melting ice, where the energy involved in the phase transition must be accounted for in heat transfer calculations.

Thermal Equilibrium

Thermal equilibrium describes the situation when all parts of a system have reached the same temperature, meaning that there is no net flow of heat between them. This concept is central to solving problems where substances at different temperatures are mixed, as it allows us to set the energy gained by cooler substances equal to the energy lost by warmer substances.

Energy Conservation in Heat Transfer

The principle of energy conservation in heat transfer states that the total energy in an isolated system remains constant. In the context of mixing substances, this means the heat lost by one part of the system must equal the heat gained by another, allowing us to establish equations that relate the specific heat capacities, masses, and temperature changes of the substances involved.

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. It is a critical concept in calorimetry because it helps quantify the thermal energy change in substances as they are heated or cooled.

Transcript

00:01 For this question, we are taking two 45 gram ice cubes and placing them into a glass containing 500 milliliters of tea.

00:12 In the previous problem, number 78, we added three ice cubes and not all of the three ice cubes melted.

00:21 However, now adding just two ice cubes, both of these ice cubes are going to melt.

00:27 We will assume, we have to assume this, even though the problem does not state it explicitly, that the two ice cubes are already at 0 degrees celsius.

00:39 We do not need to warm the ice cubes before melting them.

00:44 This is important, or it would require an additional part of the calculation.

00:50 So if the ice cubes are at 0 degrees celsius, and the t is at 20 degrees celsius, then we should be melting the ice cubes, absorbing some of the heat from the tea, and cooling the tea to somewhere between the 20 degrees celsius and the 0 degrees celsius we started at.

01:16 So the heat gained by the melting ice cubes plus the heat gained by the water associated with the melted ice cubes that warms from 0 degrees.

01:32 Celsius to whatever the final temperature is, that summed amount of energy, that summed amount of heat, should be equal to the negative of the heat lost by the t.

01:46 The negative, because the change in temperature for the water that was melted and then warming will be opposite in sign from the change in temperature for the water, the is cooling.

02:05 So that's why that negative sign is important.

02:07 We need to identify the direction of heat flow.

02:12 So what is the heat gained by the melting ice cubes? well, there are two ice cubes at 45 grams.

02:19 So we have a mass of 90 grams for the ice cubes.

02:25 If we multiply that by the heat of fusion of the ice cubes, which in this textbook is provided at 333 joules per gram, this will then be the heat that is required to melt the 90 grams of ice.

02:39 Then the ice after it has melted will warm to some final temperature based upon the heat capacity of liquid water.

02:48 That heat will be equal to the specific heat capacity of water, 4 .184 joules per gram degrees celsius, multiplied by the mass of the water that will be warming, the mass of the water will be equivalent to the mass of the ice cubes that melted, multiplied by the change in temperature, which will be the final temperature minus the initial temperature of 0 degrees celsius after the ice has melted.

03:17 We then set that equal to the negative of the heat associated with the cooling of the t, which will also be 4 .184, assuming that the t has the same specific heat capacity as water.

03:33 This was not explicit in the question, but we will make that aside.

03:37 Assumption, multiplied by the mass of the water, it tells us that we have 500 milliliters of the t...

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