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Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at -15.0\(^\circ\)C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg. Assuming no heat exchange with the surroundings, what mass of ice was added?

Short Answer

Expert verified
The added ice has a mass of approximately 604.9 g.

Step by step solution

01

Understand the Problem

We have an initial mass of water and ice in a bucket. We add more ice at a lower temperature, and it reaches thermal equilibrium with part of the water turning into ice. We are required to find out how much ice was added.
02

Identify Heat Transfer Components

There are two main processes occurring here: the ice warming up to 0°C and melting (some of it), the water cooling down and freezing, and no heat is lost to the surroundings. Therefore, heat lost by water = heat gained by the added ice.
03

Define Heat Transfer Equations

The heat required to warm the added ice from -15°C to 0°C is given by \(Q_1 = mc_i(T_f - T_i)\)where \(m\) is the added mass, \(c_i\) is the specific heat of ice (2.09 J/g°C), \(T_f = 0°C\), and \(T_i = -15°C\).
04

Calculate Heat Required to Convert Added Ice to Water

The heat required to melt the added ice is \(Q_2 = mL_f\),where \(L_f\) is the latent heat of fusion for ice, 334 J/g.
05

Set Up Heat Balance Equation

The heat lost by water cooling and freezing some part is equal to the heat gained by the added ice warming up and partially melting:\(-Q_{water} = Q_1 + Q_2\).
06

Calculate Heat Balance

First, calculate \(Q_{water}\):The mass of water initially is 1.75 kg and the final mass of ice is 0.884 kg, leaving a difference of 0.66 kg (the part of water that turned into ice). The water must cool and some must freeze. Calculating heat:\(Q_{water} = 0.66 \, \text{kg} \times (4190 \, \text{J/kg°C})(0°C - 0°C) + 0.66 \, \text{kg} \times 334 \, \text{J/g}\) for freezing. This simplifies to \(Q_{water} = 220440 \, \text{J}.\).
07

Solve for the Mass of Added Ice

Substitute and solve for an unknown mass of added ice\(-220440 = m(2.09 \times 15 + 334)\). Divide through by \(15 \, \text{g}\) to solve for \(m\): \(220440 = m(365)\), \(m = 604.9 \, \text{g}.\)
08

Verify Calculation

Check if the calculated mass satisfies equilibrium conditions for heat exchange. The mass makes sense since ice went from -15 to 0°C and melted partially, absorbing heat lost by water.

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Key Concepts

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

Thermal Equilibrium
Thermal equilibrium occurs when two or more objects at different temperatures interact with each other, exchanging heat until they reach the same temperature. In the given problem, we see this process taking place in a styrofoam bucket containing water and ice. More ice is added to this mixture, and the mixture is allowed to settle until thermal equilibrium is achieved.

During this equilibrium process, the heat gained by the ice from the water is equal to the heat lost by the water. This balance of heat, without any leak to the surroundings, means that the energy transferred is solely used to reach a uniform temperature across the system.

Understanding thermal equilibrium helps us predict how a system will behave over time if left undisturbed. In our exercise, it's the equilibrium state that reveals how much ice was needed initially. Employing this principle allows us to solve for unknown quantities based on the conservation of energy.
  • No net heat loss to surroundings
  • Temperatures of water and ice adjust until equal
  • Energy exchanged internally
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is required to raise the temperature of a substance. For instance, when ice is added to the bucket, it starts at -15°C. To increase its temperature to 0°C, energy must be absorbed. This is quantified by the specific heat capacity of ice, which is 2.09 J/g°C.

In our exercise, the specific heat capacity determines how much heat (Q_1) is needed to bring the temperature of the added ice to 0°C. It's calculated as follows: \( Q_1 = mc_i(T_f - T_i) \), where \(m\) is the mass of the ice, \(c_i\) is the specific heat capacity, and \(T_f\) and \(T_i\) are the final and initial temperatures respectively.

This concept highlights a material's resistance to temperature change as a result of heat energy input. By using specific heat capacity, we can accurately calculate the amount of energy required to change a substance’s temperature. In our case, it helps determine part of the job done by the ice in raising its temperature.
  • Ice’s specific heat: 2.09 J/g°C
  • Heat required to change temperature without phase change
  • Direct relation between heat added and temperature increase
Latent Heat of Fusion
Latent heat of fusion refers to the amount of energy needed to change a substance from solid to liquid without changing its temperature. In the task at hand, a portion of the added ice melts even as the system reaches equilibrium. This process requires the latent heat of fusion for ice, which is 334 J/g.

When calculating the energy used to melt some of the ice, we use \( Q_2 = mL_f \), where \( m \) is the mass of the ice that melts, and \( L_f \) is the latent heat of fusion. This energy calculation is vital for the added ice to metamorphose from its solid state into liquid as part of its energy transformation.

The latent heat of fusion is essential because the transformation happens at a constant temperature (0°C for ice). Thus, the entire process involves significant energetic exchanges without temperature shifts, a crucial insight for understanding phase changes in physics.
  • Heat needed for phase change: solid to liquid
  • Temperature remains constant during phase change
  • Energy calculated by mass and latent heat of fusion

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Most popular questions from this chapter

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

CP A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at 37.0\(^\circ\)C. Suppose that the water were to cool down spontaneously to form ice at 0.0\(^\circ\)C, and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20, this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

You have 750 g of water at 10.0\(^\circ\)C in a large insulated beaker. How much boiling water at 100.0\(^\circ\)C must you add to this beaker so that the final temperature of the mixture will be 75\(^\circ\)C?

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0\(^\circ\)C. A 0.250-kg block of iron at 85.0\(^\circ\)C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0\(^\circ\)F to 45.0\(^\circ\)F in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was 44.0\(^\circ\)F on January 23, 1916. The next day the temperature plummeted to -56\(^\circ\)F. What was the temperature change in Celsius degrees?

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