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

Short Answer

Expert verified
The mass of the initial water is approximately 0.492 kg.

Step by step solution

01

Calculate Heat Required to Warm Ice to 0°C

Calculate the heat needed to raise the ice temperature from -40.0°C to 0°C using the formula for heat transfer: \[ q_1 = m_{ice} \times c_{ice} \times \Delta T \]where \( q_1 \) is the heat, \( m_{ice} = 0.200 \text{ kg} \) is the mass of the ice, \( c_{ice} = 2.09 \text{ J/g°C} \) is the specific heat capacity of ice, and \( \Delta T = 40.0 \text{°C} \) is the temperature change.First, convert 0.200 kg to grams: \( 0.200 \text{ kg} = 200 \text{ g} \).\[ q_1 = 200 \text{ g} \times 2.09 \text{ J/g°C} \times 40.0 \text{°C} = 16720 \text{ J} \]
02

Calculate Heat to Melt Ice at 0°C

Calculate the heat required to melt the ice at 0°C using the formula:\[ q_2 = m_{ice} \times L_f \]where \( q_2 \) is the heat, and \( L_f = 334 \text{ J/g} \) is the latent heat of fusion for ice.\[ q_2 = 200 \text{ g} \times 334 \text{ J/g} = 66800 \text{ J} \]
03

Calculate Heat to Warm Melted Ice to Final Temperature

Calculate the heat required to warm the melted ice (now water) from 0°C to 28°C using:\[ q_3 = m_{water} \times c_{water} \times \Delta T \]where \( c_{water} = 4.18 \text{ J/g°C} \) is the specific heat capacity of water and \( \Delta T = 28.0 \text{°C} \).\[ q_3 = 200 \text{ g} \times 4.18 \text{ J/g°C} \times 28.0 \text{°C} = 23408 \text{ J} \]
04

Total Heat Required by Ice

Add up all the heat quantities:\[ q_{total} = q_1 + q_2 + q_3 \q_{total} = 16720 \text{ J} + 66800 \text{ J} + 23408 \text{ J} = 106928 \text{ J} \]
05

Calculate Heat Lost by Warm Water

The heat lost by the warm water cooling down to 28°C is calculated by:\[ q_{water} = m \times c_{water} \times \Delta T \]where \( \Delta T = 80.0 \text{°C} - 28.0 \text{°C} = 52.0 \text{°C} \).Since the total heat lost by the warm water equals the total heat gained by the ice:\[ 106928 \text{ J} = m \times 4.18 \text{ J/g°C} \times 52.0 \text{°C} \]
06

Solve for Mass of Initial Water

Isolate \( m \) in the equation:\[ m = \frac{106928 \text{ J}}{4.18 \text{ J/g°C} \times 52.0 \text{°C}} \]Calculate:\[ m \approx \frac{106928 \text{ J}}{217.36 \text{ J/g}} \approx 492 \text{ g} \]Convert grams to kilograms:\[ m = 0.492 \text{ kg} \]

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Key Concepts

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

Specific Heat Capacity
Every substance requires a certain amount of heat energy to change its temperature. The specific heat capacity is a measure of how much heat energy is needed to raise 1 gram of a substance by 1°C. It is what gives different substances their resistance to temperature change. When studying heat transfer between substances, this value is critical because it dictates how much heat is absorbed or released during a temperature change.

In our exercise, specific heat capacity plays a central role in calculating the heat exchange process. For ice, the specific heat capacity is 2.09 J/g°C. This means that each gram of ice needs 2.09 joules to raise its temperature by one degree Celsius. Similarly, for water, the value is 4.18 J/g°C, indicating the amount of energy required to heat 1 gram of water by 1°C. These differing values show why water heats up more slowly than some other substances - it requires more energy to do so.

Understanding specific heat capacity is essential:
  • It helps determine how much energy is needed to change the temperature of a substance.
  • It varies with different substances, e.g., water versus ice.
  • It's fundamental in calculating energy exchange in thermal systems.
Latent Heat of Fusion
Latent heat of fusion refers to the energy needed to change a substance from a solid to a liquid at its melting point, without a change in temperature. When ice melts into water, it absorbs heat energy but stays at 0°C until completely converted.

In our problem, to melt the 0.200 kg of ice, we need to calculate using latent heat. The latent heat of fusion for ice is given as 334 J/g. This means that each gram of ice requires 334 joules of energy to transform into water.

Latent heat of fusion is vital because:
  • It figures prominently when transitioning between solid and liquid states.
  • Without a temperature increase, it requires significant energy input.
  • It's crucial for understanding energy dynamics in phase changes.
For example, even after warming the ice to 0°C, we need additional heat energy to change its phase from solid to liquid. This concept makes latent heat significant in thermal calculations, such as predicting how much energy a particular physical transformation will need.
Temperature Change
Temperature change is an everyday experience, but scientifically, it involves understanding the energy transfer necessary to cause changes in degrees. The process by which a substance absorbs or releases heat results in a temperature increase or decrease, which is a critical aspect of heat transfer calculations.

In our exercise, the ice initially at -40°C must be warmed to 0°C. This change necessitates energy input, calculated using the specific heat capacity. Similarly, the water at 80°C is cooling down to 28°C, releasing energy in the process. Thus, temperature change isn't just a number—it explains how and why heat is exchanged.
  • It helps determine the direction of energy transfer - from hot to cold substances.
  • Plays a critical role in achieving thermal equilibrium, e.g., both substances eventually reaching 28°C.
  • Provides a basis for calculating the energy required or released during heating or cooling.
Knowing how to compute and predict temperature changes is significantly practical in designing systems and processes where thermal control is essential.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0\(^\circ\)C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0\(^\circ\)C? You can ignore the heat transferred to the container.

A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature (20.0\(^\circ\)C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 22.0\(^\circ\)C. (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

Consider a poor lost soul walking at 5 km/h on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 W, and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k'A{_s}{_k}{_i}{_n}(T{_a}{_i}{_r} - T{_s}{_k}{_i}{_n})\), where \(k'\) is 54 J/h \(\cdot\) C\(^\circ\) \(\cdot\) m\(^2\), the exposed skin area \(A{_s}{_k}{_i}{_n}\) is 1.5 m\(^2\), the air temperature \(T{_a}{_i}{_r} \)is 47\(^\circ\)C, and the skin temperature \(T{_s}{_k}{_i}{_n}\) is 36\(^\circ\)C; (iii) the skin absorbs radiant energy from the sun at a rate of 1400 W/m\(^2\); (iv) the skin absorbs radiant energy from the environment, which has temperature 47\(^\circ\)C. (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e\) = 1 and that the skin temperature is initially 36\(^\circ\)C. Which mechanism is the most important? (b) At what rate (in L/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at 36\(^\circ\)C is \(2.42 \times 10{^6}\) J/kg.) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 m\(^2\). What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.

An electric kitchen range has a total wall area of 1.40 m\(^2\) and is insulated with a layer of fiberglass 4.00 cm thick. The inside surface of the fiberglass has a temperature of 175\(^\circ\)C, and its outside surface is at 35.0\(^\circ\)C. The fiberglass has a thermal conductivity of 0.040 W /m \(\cdot\) K. (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 m\(^2\)? (b) What electric-power input to the heating element is required to maintain this temperature?

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