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Chapter 19: Problem 18

version: How long does a \(2.0 \times 10^{5} \mathrm{Btu} / \mathrm{h}\) water heater take to raise the temperature of \(40 \mathrm{gal}\) of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
3 minutes

Step by step solution

01

- Calculate the Energy Required to Heat the Water

First, calculate the energy needed to raise the temperature of the water. Use the formula: \[ Q = m \times c \times \triangle T \] Where: \( Q \) = heat energy (in Joules), \( m \) = mass of water (in kg), \( c \) = specific heat capacity of water (\( 4.186 \, \text{J/g} \, ^\text{C} \)), \( \triangle T \) = change in temperature (in 掳C). 1. Find the mass \( m \). Since 1 L of water has a mass of 1 kg, \( 150 \, \text{L} \) is equivalent to \( 150 \, \text{kg} \). 2. Calculate the change in temperature \( \triangle T \): \( 38^\text{C} - 21^\text{C} = 17^\text{C} \). 3. Plug in these values: \[ Q = 150 \, \text{kg} \times 4.186 \, \text{J/g} \, ^\text{C} \times 17^\text{C} \]
02

- Convert Energy Required to kWh

Convert the energy from Joules to kilowatt-hours (kWh): 1. Perform the multiplication: \[ Q = 150 \times 4.186 \times 17 = 10690.1 \, \text{kJ} = 10690.1 \times 1000 \, \text{J} = 10690100 \, \text{J} \] 2. Convert Joules to kWh: \( 1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J} \) \[ Q = \frac{10690100}{3.6 \times 10^6} \, \text{kWh} \, = 2.97 \, \text{kWh} \]
03

- Calculate the Time Required

Now, calculate the time needed using the heater's power rating: Use the formula: \[ \text{Time} = \frac{Q}{P} \] Where: \( Q \) = energy required in kWh, \( P \) = power of the heater (in kW). Plug in the values: \[ \text{Time} = \frac{2.97 \, \text{kWh}}{59 \, \text{kW}} \approx 0.05 \, \text{hours} \rightarrow 0.05 \times 60 \, \text{minutes} = 3 \, \text{minutes} \]
04

Conclusion

The water heater will take approximately 3 minutes to heat the water from 21掳C to 38掳C.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermodynamics. It measures the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius. For water, the specific heat capacity is quite high, around 4.186 J/g掳C. This means water can store a lot of heat energy, making it ideal for applications like cooking and heating.
Energy Conversion
Energy conversion is essential when dealing with different units in thermodynamics. In the problem, we needed to convert energy from Joules to kilowatt-hours (kWh). 1 kWh is equivalent to 3.6 x 10^6 Joules. This conversion is crucial for calculating how long a heater needs to run, given its power rating in kilowatts (kW). Always remember to use the correct conversion factors to ensure accurate results.
Temperature Change Calculation
Temperature change calculation involves finding the difference between the final and initial temperatures. In our problem, the water temperature changed from 21掳C to 38掳C. So the change in temperature (螖T) is 38掳C - 21掳C = 17掳C. This value helps in determining the amount of heat energy required to achieve this temperature change. Always make sure to use consistent units for temperatures when performing these calculations.
Heat Energy Formula
The heat energy formula helps calculate the energy required to heat a substance. The formula is Q = m 脳 c 脳 螖T, where Q is the heat energy in Joules, m is the mass in kilograms, c is the specific heat capacity, and 螖T is the temperature change in Celsius. For water, with a mass of 150 kg, a specific heat capacity of 4.186 J/g掳C, and a temperature change of 17掳C, the energy required is calculated as follows: Q = 150 kg 脳 4.186 J/g掳C 脳 17掳C. Properly applying this formula is key to solving many thermodynamics problems.

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Most popular questions from this chapter

A small electric immersion heater is used to heat \(100 \mathrm{~g}\) of water for a cup of instant coffee. The heater is labeled "200 watts," so it converts electrical energy to thermal energy that is transferred to the water at this rate. Calculate the time required to bring the water from \(23^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) ignoring any thermal energy that transfers out of the cup.

A tank of water has been outdoors in cold weather, and a slab of ice \(5.0 \mathrm{~cm}\) thick has formed on its surface (Fig. 19-42). The air above the ice is at \(-10^{\circ} \mathrm{C}\). Calculate the rate of formation of ice (in centimeters per hour) on the ice slab. Take the thermal conductivity and density of ice to be \(0.0040 \mathrm{cal} / \mathrm{s} \cdot \mathrm{cm} \cdot \mathrm{C}^{\circ}\) and \(0.92 \mathrm{~g} / \mathrm{cm}^{3}\). Assume that energy is not transferred through the walls or bottom of the tank.

Converting kinetic energy into thermal energy produces small rises in temperature. This was in part responsible for the difficulty in discovering the law of conservation of energy. It also implies that hot objects contain a lot of energy. (This latter comment is largely responsible for the industrial revolution in the 19 th century.) To get some feel for these numbers, assume all mechanical energy is converted to thermal energy and carry out three estimates: (a) A steel ball is dropped from a height of \(3 \mathrm{~m}\) onto a concrete floor. It bounces a large number of times but eventually comes to rest. Estimate the ball's rise in temperature. (b) Suppose the steel ball you used in part (a) is at room temperature. If you converted all its thermal energy to translational kinetic energy, how fast would it be moving? (Give your answer in units of miles per hour. Also, ignore the fact that you would have to create momentum.) (c) Suppose a nickel-iron meteor falls to Earth from deep space. Estimate how much its temperature would rise on impact.

At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius and (b) half that of the Celsius?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\) and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) How would the answer to (a) change if, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} ?\)
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