/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 mmh A 0.35-kg coffee mug is made... [FREE SOLUTION] | 91Ó°ÊÓ

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mmh A 0.35-kg coffee mug is made from a material that has a specific heat capacity of 920 \({J} /({kg} \cdot {C}^{\circ})\) and contains 0.25 kg of water. The cup and water are at 15 C. To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

Short Answer

Expert verified
The minimum power rating of the heater is approximately 646 W.

Step by step solution

01

Identify Given Values

Start by identifying the values given in the problem:- Mass of the coffee mug, \(m_{mug} = 0.35\, kg\)- Specific heat capacity of the coffee mug, \(c_{mug} = 920\, J/(kg \cdot C^{\circ})\)- Mass of water, \(m_{water} = 0.25\, kg\)- Specific heat capacity of water, \(c_{water} = 4186\, J/(kg \cdot C^{\circ})\)- Initial temperature, \(T_i = 15^{\circ}C\)- Final temperature, \(T_f = 100^{\circ}C\)- Time to boil, \(time = 3\, minutes = 180\, seconds\)
02

Calculate Temperature Change

The change in temperature \(\Delta T\) for both the water and the mug is the same:\[\Delta T = T_f - T_i = 100^{\circ}C - 15^{\circ}C = 85^{\circ}C\]
03

Calculate Heat Required for Water

The heat required to raise the temperature of the water can be calculated using:\[Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T\]Substitute the given values:\[Q_{water} = 0.25 \times 4186 \times 85 = 88842.5\, J\]
04

Calculate Heat Required for Coffee Mug

The heat required to raise the temperature of the mug is:\[Q_{mug} = m_{mug} \cdot c_{mug} \cdot \Delta T\]Substitute the given values:\[Q_{mug} = 0.35 \times 920 \times 85 = 27370\, J\]
05

Calculate Total Heat Required

The total heat energy required is the sum of the heat for the mug and the water:\[Q_{total} = Q_{water} + Q_{mug} = 88842.5 + 27370 = 116212.5\, J\]
06

Calculate Minimum Power Rating of Heater

Power is the rate of energy transfer with time:\[P = \frac{Q_{total}}{time}\]Substitute the total heat and the time:\[P = \frac{116212.5}{180} \approx 645.6\, W\]
07

Conclude Calculation

Hence, the minimum power rating of the heater is approximately 646 watts.

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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 property that tells us how much heat energy is required to raise the temperature of a certain mass of a substance by one degree Celsius. In this exercise, we dealt with two specific values for the substances involved:

  • Coffee mug: 920 \( J/(kg \cdot C^{\circ}) \)
  • Water: 4186 \( J/(kg \cdot C^{\circ}) \)
The large specific heat capacity of water means it requires more energy to heat compared to the material of the coffee mug. Knowing this helps us calculate the energy required to bring both the mug and water to a desired temperature.
The concept of specific heat capacity is vital in thermodynamics as it determines how substances react to energy changes.
Energy Transfer
Energy transfer in this context refers to the process of transferring heat from the heater to both the coffee mug and water. The formula used to determine this transfer is:\[ Q = m \cdot c \cdot \Delta T \]Where:

  • \( Q \) is the heat energy transferred
  • \( m \) is the mass
  • \( c \) is the specific heat capacity
  • \( \Delta T \) is the change in temperature
This equation allows us to find out how much heat energy is specifically required to bring the mug and water from 15°C to 100°C. By carefully calculating \( Q \) for both the water and the mug independently, and then summing them up, we get the total energy required for the entire heating process.
Power Calculation
Calculating power involves determining how quickly energy is transferred or used. Power is given by the formula:\[ P = \frac{Q_{total}}{time} \]Here:

  • \( P \) is the power
  • \( Q_{total} \) is the total energy required
  • \( time \) is the duration of energy transfer
In this exercise, the total energy needed to bring the water and mug to a boil is divided by the time taken (180 seconds). The calculated power helps us determine the minimum power rating of the heater required to achieve this within the given timeframe. The real-world application of this concept allows us to design appliances like heaters efficiently.
Temperature Change
Temperature change is a crucial aspect of solving problems in thermodynamics. In this exercise, the change in temperature \( \Delta T \) was calculated as the final temperature minus the initial temperature:\[ \Delta T = T_f - T_i = 100^{\circ}C - 15^{\circ}C = 85^{\circ}C \]This means that both the coffee mug and the water need to be heated by 85°C.

The change in temperature is directly related to the amount of energy required to heat an object. Knowing the specific starting and ending temperatures allows us to precisely determine the energy demands of the process. This fundamental calculation underpins many practical applications, ensuring we use energy efficiently and effectively.

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Most popular questions from this chapter

A snow maker at a resort pumps 130 kg of lake water per minute and sprays it into the air above a ski run. The water droplets freeze in the air and fall to the ground, forming a layer of snow. If all the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of \(-7.0^{\circ} {C},\) how much heat does the snow-making process release each minute? Assume that the temperature of the lake water is \(12.0^{\circ} \mathrm{C},\) and use \(2.00 \times 10^{3} {J} /({kg} \cdot {C}^{\circ})\) for the specific heat capacity of snow.

The latent heat of vaporization of \({H}_{2} {O}\) at body temperature \((37.0^{\circ} {C})\) is \(2.42 \times 10^{6} {J} / {kg} .\) To cool the body of a \(75-{kg}\) jogger [average specific heat capacity \(=3500 {J} /({kg} \cdot {C}^{9}) ]\) by \(1.5 {C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

If a nonhuman civilization were to develop on Saturn’s largest moon, Titan, its scientists might well devise a temperature scale based on the properties of methane, which is much more abundant on the surface than water is. Methane freezes at \(-182.6^{\circ} {C}\) on Titan, and boils at \(-155.2^{\circ} {C}\) Taking the boiling point of methane as \(100.0^{\circ} {M}\) (degrees Methane) and its freezing point as \(0^{\circ} {M},\) what temperature on the Methane scale corresponds to the absolute zero point of the Kelvin scale?

ssm At a fabrication plant, a hot metal forging has a mass of 75 \({kg}\) and a specific heat capacity of 430 \({J} / {kg} \cdot {C}^{\circ}\) . To harden it, the forging is immersed in 710 \({kg}\) of oil that has a temperature of \(32^{\circ} {C}\) and a specific heat capacity of 2700 \({J} / {kg} \cdot {C}^{\circ}\) ). The final temperature of the oil and forging at thermal equilibrium is \(47^{\circ} {C}\) . Assuming that heat flows only between the forging and the oil, determine the initial temperature of the forging.

Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} {C}\) . When the temperature drops to \(-14^{\circ} {C}\) , the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} {N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross- sectional area of \(1.6 \times 10^{-5} {m}^{2},\) and it is made from a material whose coefficient of linear expansion is $$2.5 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}$$ . What is Young's modulus for the material from which the ruler is made?

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