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An insulated beaker with negligible mass contains \(0.250 \mathrm{~kg}\) of water at \(75.0^{\circ} \mathrm{C}\). How many kilograms of ice at \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(40.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The mass of ice needed is the solution obtained from the equation in Step 7.

Step by step solution

01

Identify knowns and unknowns

The mass of the water is \(0.250 \mathrm{~kg}\), its initial temperature is \(75.0^{\circ} \mathrm{C}\), and the final temperature is \(40.0^{\circ} \mathrm{C}\). The initial temperature of the ice is \(-20.0^{\circ} \mathrm{C}\) and it ends in the same final temperature. The unknown is the mass of the ice, denoted as \(m_{\text{ice}}\).
02

Calculate heat gained by ice

First, ice will warm up from \(-20.0^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) without changing the phase. This heat can be calculated by using the equation: \(Q_{1} = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T_{1}\), where \(c_{\text{ice}} = 2.09 \mathrm{~kJ/kg\cdot K}\) is the specific heat of ice and \(\Delta T_{1} = 0 - (-20) = 20 \mathrm{~K}\) is the change in temperature.
03

Calculate heat gained during phase change

When ice reaches \(0^{\circ} \mathrm{C}\), it will start melting, which requires additional heat called heat of fusion. This can be calculated by using the formula: \(Q_{2} = m_{\text{ice}} \cdot L_{F}\), where \(L_{F} = 334 \mathrm{~kJ/kg}\) is the latent heat of fusion of ice.
04

Calculate heat gained by the melted ice

Once all the ice has melted, the resulting water will continue to heat up from \(0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\). The heat gain can be calculated by using the equation: \(Q_{3} = m_{\text{ice}} \cdot c_{\text{water}} \cdot \Delta T_{2}\), where \(c_{\text{water}} = 4.18 \mathrm{~kJ/kg\cdot K}\) is the specific heat of water and \(\Delta T_{2} = 40 - 0 = 40 \mathrm{~K}\) is the temperature change.
05

Calculate heat lost by the water

The heat lost by the water as it cools from \(75.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\) can be calculated using the equation: \(Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{3}\), where \(m_{\text{water}} = 0.250 \mathrm{~kg}\) is the mass of the water and \(\Delta T_{3} = 40 - 75 = -35 \mathrm{~K}\) is the temperature change.
06

Apply conservation of energy

The total heat gained by the ice must be equal to the total heat lost by the water. Therefore: \(Q_{1} + Q_{2} + Q_{3} = Q_{\text{water}}\). Simplifying the expression will lead to the equation to calculate the mass of the ice.
07

Solve for unknown

Solving the equation from Step 6 will give the unknown mass of the ice, \(m_{\text{ice}}\).

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Key Concepts

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

Calorimetry
Calorimetry is the science of measuring the heat of chemical reactions or physical changes as well as heat capacity. Here, you can think of a calorimeter as a kind of device used to measure amounts of heat exchanged or transformed in a system. Let’s understand the exercise: an insulated beaker can be seen as a simple calorimeter where the water and ice exchange heat without any loss to the surrounding. The main idea in calorimetry is to ensure that all the heat exchange happens between the substances inside the calorimeter.

In our problem, the exercise uses the calorimetry principle to discover how much ice is needed to reduce the hot water's temperature to a specified amount. By keeping track of all the temperature changes and phase transitions, we can maintain a balance of heat exchange, ensuring that the heat lost by the water is equal to the heat gained by the ice.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Each material has its own specific heat capacity, meaning they require different amounts of heat to increase their temperature. This property is crucial in our exercise.

We observe two different specific heat capacities in action: one for ice and one for water. For instance:
  • The specific heat capacity of ice: 2.09 kJ/kg·K
  • The specific heat capacity of water: 4.18 kJ/kg·K
When ice is warming up from (-20.0^{ ext{C}}) to (0^{ ext{C}}), we multiply by the ice's specific heat. When the melted ice (now liquid water) warms up to (40^{ ext{C}}), we use the specific heat capacity of water. This calculation helps us understand exactly how much energy is absorbed at each stage.
Latent Heat
Latent heat is the heat required to change the phase of a substance, like melting or freezing, without changing its temperature. It is an essential concept when dealing with phase changes as it involves a large amount of energy, even though the temperature does not change during the transition.

In the exercise, when the ice reaches (0^{ ext{C}}), it needs the latent heat of fusion to turn into water. This is calculated as:
  • The latent heat of fusion for ice: 334 kJ/kg
This tells us how much energy is required to convert 1 kg of ice at 0°C to water at the same temperature. The process consumes a lot of heat, contributing significantly to reducing the overall temperature of the system.
Conservation of Energy
Conservation of energy is a fundamental concept stating that energy cannot be created or destroyed in an isolated system. Only converted from one form to another. In our context, this is pivotal in letting us set up the calculation that solves our problem.

In the insulated beaker, the heat lost by the hot water is not escaping to the environment; it’s being transferred to the ice. Therefore, the heat gained by the ice must exactly equal the heat lost by the water, representing a closed and balanced system. Mathematically, this is expressed in the exercise as:

(Q_{1} + Q_{2} + Q_{3} = Q_{ ext{water}}). This equation ensures everything is accounted for, providing a way to solve for the unknown mass of ice needed to achieve the final equilibrium temperature.

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Most popular questions from this chapter

A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{~m} \times 0.95 \mathrm{~m} \times 5.0 \mathrm{~cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8 \mathrm{~cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C},\) and the outside air temperature is \(-8.0^{\circ} \mathrm{C}\). (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window \(0.500 \mathrm{~m}\) on a side is inserted in the door? The glass is \(0.450 \mathrm{~cm}\) thick, and the glass has a thermal conductivity of \(0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional \(12.0 \mathrm{~cm}\) of glass.

In a container of negligible mass, \(0.200 \mathrm{~kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\). No heat is lost to the surroundings. If the final temperature of the system is \(28.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C}\) ?

Hot Air in a Physics Lecture. (a) A typical student listening attentively to a physics lecture has a heat output of \(100 \mathrm{~W}\). How much heat energy does a class of 140 physics students release into a lecture hall over the course of a 50 min lecture? (b) Assume that all the heat energy in part (a) is transferred to the \(3200 \mathrm{~m}^{3}\) of air in the room. The air has specific heat \(1020 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and density \(1.20 \mathrm{~kg} / \mathrm{m}^{3} .\) If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50 min lecture? (c) If the class is taking an exam, the heat output per student rises to \(280 \mathrm{~W}\). What is the temperature rise during \(50 \mathrm{~min}\) in this case?

An asteroid with a diameter of \(10 \mathrm{~km}\) and a mass of \(2.60 \times 10^{15} \mathrm{~kg}\) impacts the earth at a speed of \(32.0 \mathrm{~km} / \mathrm{s},\) landing in the Pacific Ocean. If \(1.00 \%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(\left.10.0^{\circ} \mathrm{C}\right),\) what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{~kg} .\)

A industrious explorer of the polar regions has devised a contraption for melting ice. It consists of a sealed \(10 \mathrm{~L}\) cylindrical tank with a porous grate separating the top half from the bottom half. The bottom half includes a paddle wheel attached to an axle that passes outside the cylinder, where it is attached by a gearbox and pulley system to a stationary bicycle. Pedaling the bicycle rotates the paddle wheel inside the cylinder. The tank includes \(6.00 \mathrm{~L}\) of water and \(3.00 \mathrm{~kg}\) of ice at \(0.0^{\circ} \mathrm{C}\). The water fills the bottom chamber, where it may be agitated by the paddle wheel, and partially fills the upper chamber, which also includes the ice. The bicycle is pedaled with an average torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) at a rate of 30.0 revolutions per minute. The system is \(70 \%\) efficient. (a) For what length of time must the explorer pedal the bicycle to melt all the ice? (b) How much longer must he pedal to raise the temperature of the water to \(10.5^{\circ} \mathrm{C}\) ?

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