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Chapter 17: Problem 59

An insulated beaker with negligible mass contains 0.250 \(\mathrm{kg}\) of water at a temperature of \(75.0^{\circ} \mathrm{C}\) . How many kilograms of ice at a temperature of \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(30.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Approximately 0.0945 kg of ice is needed.

Step by step solution

01

Identify the energy change in the water

The water initially has a mass of 0.250 kg and a temperature of 75.0掳C. It cools down to 30.0掳C. We calculate the energy lost by the water using the specific heat capacity of water, which is 4,186 J/kg掳C. \[Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T\]\[Q_{water} = 0.250 \times 4186 \times (30.0 - 75.0)\]\[ Q_{water} = 0.250 \times 4186 \times (-45) = -47,092.5 \, \text{J} \]
02

Calculate the energy required to warm the ice to 0掳C

The ice, initially at -20.0掳C, must first be raised to 0掳C. We use the specific heat capacity of ice, which is 2,090 J/kg掳C.\[Q_{1} = m_{ice} \cdot c_{ice} \cdot \Delta T\]\[Q_{1} = m_{ice} \times 2090 \times (0 - (-20))\]\[Q_{1} = m_{ice} \times 2090 \times 20 = 41800 \, m_{ice} \, \text{J} \]
03

Calculate the energy required to melt the ice

After the ice reaches 0掳C, it needs to be melted. The latent heat of fusion for ice is 334,000 J/kg.\[Q_{2} = m_{ice} \cdot L_f\]\[Q_{2} = m_{ice} \times 334,000\]
04

Calculate the energy required to warm the melted ice to 30掳C

Once the ice has melted, the resulting water needs to be warmed from 0掳C to 30掳C. We use the specific heat capacity of water (4,186 J/kg掳C) for this calculation.\[Q_{3} = m_{water} \cdot c_{water} \cdot \Delta T\]\[Q_{3} = m_{ice} \times 4186 \times 30\]
05

Set up the energy balance equation

The total heat gained by the ice should equal the heat lost by the water. Thus, \[Q_{1} + Q_{2} + Q_{3} = -Q_{water}\]Substituting the expressions for \(Q_{1}\), \(Q_{2}\), and \(Q_{3}\) from steps 2, 3, and 4:\[41800 \times m_{ice} + 334000 \times m_{ice} + 125580 \times m_{ice} = 47092.5\]
06

Solve for the mass of the ice

Combine and simplify the equation from step 5:\[41800 \times m_{ice} + 334000 \times m_{ice} + 125580 \times m_{ice} = 47092.5\]Combine like terms:\[498380 \times m_{ice} = 47092.5\]Solve for \(m_{ice}\):\[m_{ice} = \frac{47092.5}{498380} \approx 0.0945 \, \text{kg} \]

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Key Concepts

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

Specific Heat Capacity
The specific heat capacity is an essential concept in thermodynamics. It refers to the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Celsius. This property is crucial for understanding how different materials react to changes in temperature. In this exercise, the specific heat capacity of water is important because it helps us calculate how much energy is lost when water cools down from 75掳C to 30掳C.

For water, this value is 4,186 J/kg掳C. The formula to find the energy change due to temperature change is: \( Q = m \cdot c \cdot \Delta T \), where:
  • \( Q \) is the heat energy in joules (J)
  • \( m \) is the mass in kilograms (kg)
  • \( c \) is the specific heat capacity (J/kg掳C)
  • \( \Delta T \) is the change in temperature in degrees Celsius (掳C)
This formula allows us to calculate how much energy the 0.250 kg of water at 75掳C releases as it cools to 30掳C.
Latent Heat of Fusion
The latent heat of fusion is the energy required to change a solid into a liquid without changing its temperature. It's a critical concept for situations where a substance transitions between solid and liquid states, such as in this exercise where ice melts into water.

For ice, the latent heat of fusion is 334,000 J/kg. This means it takes a substantial amount of energy to transform ice at 0掳C into water at the same temperature. When calculating the energy needed to melt ice, the formula used is: \( Q = m \cdot L_f \), where:
  • \( Q \) is the heat energy in joules (J)
  • \( m \) is the mass of the ice (kg)
  • \( L_f \) is the latent heat of fusion for the substance (J/kg)
No temperature change occurs during this process, highlighting the importance of understanding latent heat when involving phase changes.
Energy Balance Equation
The energy balance equation is fundamental in thermodynamics, ensuring the conservation of energy within a system. It dictates that the total energy gained by one part of the system must be equal to the energy lost by another part.

In our example, the energy lost by the water is gained by the ice. This means the cooling of water and the warming, melting, and further heating of ice must exactly exchange the same amount of energy. The equation is expressed as:\[ Q_{1} + Q_{2} + Q_{3} = -Q_{water} \]where:
  • \( Q_{1} \) is the energy to warm the ice to 0掳C
  • \( Q_{2} \) is the energy to melt the ice
  • \( Q_{3} \) is the energy to warm the melted ice to 30掳C
  • \( -Q_{water} \) is the energy lost by the water as it cools
Understanding this balance lets us solve for unknowns, such as the mass of ice needed to achieve our desired final system temperature.

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Most popular questions from this chapter

A carpenter builds an exterior house wall with a layer of wood 3.0 \(\mathrm{cm}\) thick on the outside and a layer of Styrofoam insulation 2.2 \(\mathrm{cm}\) thick on the inside wall surface. The wood has \(k=0.080 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) , and the Styrofoam has \(k=0.010 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The interior surface temperature is \(19.0^{\circ} \mathrm{C}\) , and the exterior surface temperature is \(-10.0^{\circ} \mathrm{C}\) (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

A copper calorimeter can with mass 0.100 kg contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Two beakers of water, \(A\) and \(B\) , initially are at the same temperature. The temperature of the water in beaker \(A\) is increased \(10 F^{\circ},\) and the temperature of the water in beaker \(B\) is increased 10 \(\mathrm{K}\) . After these temperature changes, which beaker of water has the higher temperature? Explain.

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 \(10.0^{\circ} \mathrm{C} )\) , what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{kg}\) .)

Food Intake of a Hamster. The energy output of an animal engaged in an activity is called the basal metabolic rate (BMR) and is a measure of the conversion of food energy into other forms of energy. A simple calorimeter to measure the BMR consists of an insulated box with a thermometer to measure the temperature of the air. The air has density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and specific heat 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . A 50.0 -g hamster is placed in a calorimeter that contains 0.0500 \(\mathrm{m}^{3}\) of air at room temperature. (a) When the hamster is running in a wheel, the temperature of the air in the calorimeter rises 1.60 \(\mathrm{C}^{\circ}\) per hour. How much heat does the running hamster generate in an hour? Assume that all this heat goes into the air in the calorimeter. You can ignore the heat that goes into the walls of the box and into the thermometer, and assume that no heat is lost to the surroundings. (b) Assuming that the hamster converts seed into heat with an efficiency of 10\(\%\) and that hamster seed has a food energy value of 24 \(\mathrm{J} / \mathrm{g}\) , how many grams of seed must the hamster eat per hour to supply this energy?
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