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Chapter 14: Problem 20

\(\cdot\) In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200.0 \(\mathrm{W}\) electric immersion heater in 0.320 \(\mathrm{kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required if all of the heater's power goes into heating the water?

Short Answer

Expert verified
The heat required is 80,371.2 J, and it takes about 6.70 minutes to heat the water.

Step by step solution

01

Understanding the Problem

We'll divide the problem into two parts. First, calculate the heat needed to increase the water's temperature. Then, calculate the time required for the heater to provide this amount of energy.
02

Calculate the Heat Energy Required

We need to calculate the heat energy using the formula: \[ q = mc\Delta T \] where:- \( m = 0.320 \, \text{kg} \)- \( c \) is the specific heat capacity of water, approximately \( 4.186 \, \text{J/g}^\circ\text{C} \) or \( 4186 \, \text{J/kg}^\circ\text{C} \)- \( \Delta T = 80.0^{\circ}\text{C} - 20.0^{\circ}\text{C} = 60.0^{\circ}\text{C} \)Substitute these values into the formula:\[ q = 0.320 \, \text{kg} \times 4186 \, \text{J/kg}^\circ\text{C} \times 60.0^{\circ}\text{C} = 80,371.2 \, \text{J} \]
03

Calculate the Time Required

We use the formula for power: \[ P = \frac{q}{t} \] Rearranging for time \( t \):\[ t = \frac{q}{P} \]Substitute the values \( q = 80,371.2 \, \text{J} \) and \( P = 200.0 \, \text{W} \):\[ t = \frac{80,371.2 \, \text{J}}{200.0 \, \text{W}} = 401.856 \, \text{s} \]
04

Converting Seconds to Minutes

Since the result is in seconds, we'll convert it to minutes for practicality:\[ t = \frac{401.856 \, \text{s}}{60 \, \text{s/min}} \approx 6.70 \, \text{minutes} \]

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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 measure of how much thermal energy is required to raise the temperature of a given mass of a substance by one degree Celsius. Not all substances heat up at the same rate.
This is why specific heat capacity is crucial in understanding how different materials react to thermal energy. Imagine holding a metal spoon and a wooden spoon over a flame.
The metal spoon heats up faster than the wooden spoon.
This difference is due to their specific heat capacities. Water has a relatively high specific heat capacity, approximately 4186 J/kg掳C, meaning it takes a lot of energy to change its temperature.
  • Units: Joules per kilogram per degree Celsius (J/kg掳C)
  • Varies with material
  • Important for designing heating and cooling systems, like the coffee heater example
Electric Power
Electric power is the rate at which electrical energy is transferred by an electric circuit.
It is measured in watts (W).
One watt corresponds to the consumption of one joule of energy per second. In the context of the heated cup of coffee, the immersion heater utilizes electric power to convert electrical energy into thermal energy. This conversion is key to increasing the water's temperature. Electric power is given by the formula:\[ P = \frac{E}{t} \]Where:
  • \(P\) is power (in watts)
  • \(E\) is energy (in joules)
  • \(t\) is time (in seconds)
This means a 200 W heater can provide 200 Joules of energy every second, efficiently warming up water for our student鈥檚 coffee.
Thermal Energy
Thermal energy is the total internal kinetic energy of particles in a substance. When you heat water to make coffee, you're transferring thermal energy from the heater to the water molecules. The more thermal energy is transferred, the more the particles vibrate, increasing the water's temperature. In practical scenarios like boiling water, understanding thermal energy helps in predicting how long a process will take or how much heat energy is required.
Mathematically, this is represented as:\[ q = mc\Delta T \]Where:
  • \(q\) is the thermal energy (in joules)
  • \(m\) is the mass of the water (in kilograms)
  • \(c\) is the specific heat capacity
  • \(\Delta T\) is the change in temperature (in 掳C)
This calculation is vital when you need to know how much energy your immersion heater should provide.
Temperature Change
Temperature change signifies the difference between the initial and final temperatures of a substance when energy is added or removed. It is a critical factor in calculating the thermal energy needed and in assessing how quickly a process like heating water can happen. In our coffee example, the water's temperature is raised from 20掳C to 80掳C, giving a temperature change \(\Delta T\) of 60掳C. Understanding temperature change helps in:
  • Designing heating systems
  • Predicting energy consumption
  • Adjusting heat transfer according to needs
Whether you're a student attempting an all-nighter or designing industrial systems, grasping temperature change ensures effective thermal management. Combining this understanding with specific heat capacity forms the basis for effective problem-solving in physics.

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Most popular questions from this chapter

Temperatures in biomedicine. (a) Normal body temperature. The average normal body temperature measured in the mouth is 310 \(\mathrm{K}\) . What would Celsius and Fahrenheit thermometers read for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\) . What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about 7 \(\mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

\(\bullet\) A \(25,000\) -kg subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\)

A copper calorimeter can with mass 0.100 \(\mathrm{kg}\) contains 0.160 \(\mathrm{kg}\) of water and 0.018 \(\mathrm{kg}\) 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 can, what is the final temperature of the system if no heat is lost to the surroundings?

\(\bullet\) Burning fat by exercise. Each pound of fat contains 3500 food calories. When the body metabolizes food, 80\(\%\) of this energy goes to heat. Suppose you decide to run without stopping, an activity that produces 1290 \(\mathrm{W}\) of metabolic power for a typical person. (a) For how many hours must you run to burn up 1 lb of fat? Is this a realistic exercise plan? (b) If you followed your planned exercise program, how much heat would your body produce when you burn up a pound of fat (c) If you needed to get rid of all of this excess heat by evaporating water (i.e., sweating), how many liters would you need to evaporate? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\)

. A thermos for liquid helium. A physicist uses a cylindrical metal can 0.250 \(\mathrm{m}\) high and 0.090 \(\mathrm{m}\) in diameter to store liquid helium at \(4.22 \mathrm{K} ;\) at that temperature the heat of vaporization of helium is \(2.09 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, 77.3 \(\mathrm{K}\) , with vacuum between the can and the surrounding walls. How much helium is lost per hour? The emissivity of the metal can is 0.200 . The only heat transfer between the metal can and the surrounding walls is by radiation.
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