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Chapter 17: Problem 48

An ice-cule tray of negligible mass contains 0.350 \(\mathrm{kg}\) of water at \(18.0^{\circ} \mathrm{C} .\) How much heat must be removed to cool the water to \(0.00^{\circ} \mathrm{C}\) and freeze it? Express your answer in joules, calories, and Btu.

Short Answer

Expert verified
The total heat removed is approximately 143343 J, 34253 cal, and 136 BTU.

Step by step solution

01

Determine the Heat to Cool Water to Freezing Point

First, calculate the amount of heat required to cool the water from its initial temperature to its freezing point, which is 0掳C. The specific heat capacity of water is. The formula to calculate heat is: \[ Q_1 = m imes c imes riangle T \]where \( m = 0.350 \text{ kg} \), \( c = 4.186 \text{ J/g掳C} \), and \( \triangle T = 18.0掳C - 0掳C \). Note that 1 kg = 1000 g. So, substitute these values in:\[ Q_1 = 350 \times 4.186 \times 18.0 = 26442.6 \text{ J} \]
02

Determine the Heat to Freeze the Water

Now calculate the heat required to freeze the water. The latent heat of fusion for water is 334 J/g. Use the formula:\[ Q_2 = m \times L_f \]where \( L_f = 334 \text{ J/g} \). Substitute the values into the equation:\[ Q_2 = 350 \times 334 = 116900 \text{ J} \]
03

Calculate the Total Heat Removed

Add the heat needed to cool the water to 0掳C (\( Q_1 \)) and the heat required to freeze the water (\( Q_2 \)) to find the total heat removed:\[ Q_{total} = Q_1 + Q_2 = 26442.6 + 116900 = 143342.6 \text{ J} \]
04

Convert Joules to Calories

To convert the total heat from joules to calories, use the conversion factor: 1 calorie = 4.184 joules. Therefore:\[ Q_{total} = \frac{143342.6}{4.184} \approx 34252.7 \text{ cal} \]
05

Convert Joules to BTU

To convert joules to British Thermal Units (BTU), use the conversion factor: 1 BTU = 1055.06 joules. Thus:\[ Q_{total} = \frac{143342.6}{1055.06} \approx 135.9 \text{ BTU} \]

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Key Concepts

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

Understanding Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermodynamics that represents the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is quite significant at 4.186 J/g掳C. This means that water can absorb a lot of heat before its temperature changes.
For practical purposes, the specific heat capacity tells us how much energy we need to add or remove to change the temperature of a given mass of a substance. In the exercise we dealt with, the specific heat capacity of water helps calculate how much heat is needed to lower the water's temperature from 18.0掳C to 0.0掳C.
  • The formula used is: \[ Q = m \times c \times \Delta T \] Here, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
Through this relationship, the energy required to bring the water to its freezing point was found to be 26442.6 J, showcasing the high capacity for water to store thermal energy.
The Concept of Latent Heat of Fusion
Latent heat of fusion is another critical thermodynamic concept that explains the energy required to change a substance from solid to liquid or vice versa at constant temperature. For water, the latent heat of fusion is 334 J/g, which means that 334 joules of energy are needed to melt or freeze one gram of ice without changing its temperature.
In our exercise, after cooling the water to 0掳C, the next step is freezing it. The heat removed during this phase change does not decrease the temperature further but only changes the state from liquid to solid.
  • The formula to determine the heat for the phase change: \[ Q = m \times L_f \] \(L_f\) is the latent heat of fusion.
To complete the phase change for the example problem, 116900 J of heat was removed, allowing the water to transition into ice perfectly.
Exploring Heat Energy Conversion
Heat energy conversion is essential for understanding various energy units and making useful comparisons between them. In our exercise, once the total heat was found, converting it into different units allowed for universal understanding and comparison.
Firstly, converting joules to calories, where the relationship is 1 calorie = 4.184 joules, allows us to express heat in a more human-scale unit. This conversion showed that we needed about 34252.7 calories for our task.
Additionally, converting to British Thermal Units (BTU) is useful, especially in regions and industries where BTU is a standard unit for measuring heat. This involves knowing that 1 BTU = 1055.06 joules, translating our example's required energy into approximately 135.9 BTU.
  • Formulas for conversion include:
    \[ \text{calories} = \frac{\text{joules}}{4.184} \]\ \[ \text{BTU} = \frac{\text{joules}}{1055.06} \]
These conversions allow for diverse applications and better communication of thermal energy across different scientific and engineering fields.

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Most popular questions from this chapter

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0-kg man to cool his body 1.00 \(\mathrm{C}^{\circ} ?\) The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.31\() .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{cm}^{3}\right)\).

Treatment for a Stroke. One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\) . To treat a 70.0 -kg patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C}\) ) you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat of the human body is \(3480 \mathrm{J} / \mathrm{kg} \cdot \mathrm{C}^{\circ},\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C} .\)

The emissivity of tungsten is \(0.350 .\) A tungsten sphere with radius 1.50 \(\mathrm{cm}\) is suspended within a large evacuated enclosure whose walls are at 290.0 \(\mathrm{K}\) . What power input is required to maintain the sphere at a temperature of 3000.0 \(\mathrm{K}\) if heat conduction along the supports is neglected?

In a container of negligible mass, 0.0400 kg of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

On-Demand Water Heaters. Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is 2.5 gal/min (9.46 L/min) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?
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