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(II) A small immersion heater is rated at \(350 \mathrm{~W}\). Estimate how long it will take to heat a cup of soup (assume this is \(250 \mathrm{~mL}\) of water) from \(15^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\).

Short Answer

Expert verified
It takes approximately 179 seconds to heat the soup.

Step by step solution

01

Convert Volume to Mass

Since the density of water is approximately 1 g/mL, the mass of the water is equivalent to its volume. Therefore, for 250 mL of water, the mass is 250 grams or 0.25 kg.
02

Calculate Temperature Change

The temperature change \(\Delta T\)is the final temperature minus the initial temperature. For this problem, \(\Delta T = 75^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 60^{\circ} \mathrm{C}\).
03

Energy Required to Heat the Water

Using the formula \(Q = mc\Delta T\),where \(m\)is the mass, \(c = 4186\, \mathrm{J/kg\cdot^{\circ}C}\)is the specific heat capacity of water, and \(\Delta T\)is the temperature change: \[Q = 0.25\, \mathrm{kg} \times 4186\, \mathrm{J/kg\cdot^{\circ}C} \times 60\,^{\circ} \mathrm{C} = 62790\, \mathrm{J}\].
04

Determine Time Using Power

Power is energy per unit time, given by \(P = \frac{Q}{t}\),where \(P = 350\, \mathrm{W}\)(equivalent to 350 J/s). Solving for time \(t\),we get \[t = \frac{Q}{P} = \frac{62790\, \mathrm{J}}{350\, \mathrm{W}} \approx 179.4\, \mathrm{s}\].

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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 fascinating concept in thermodynamics. It represents the amount of heat energy required to raise the temperature of a unit mass of a substance by 1 degree Celsius. The specific heat capacity varies between different substances. For water, which is commonly used in thermodynamics problems due to its prevalence and distinct properties, the specific heat capacity is particularly high, valued at approximately 4186 J/kg°C. This means water requires a significant amount of energy to heat up.
Water's high specific heat capacity has practical applications, such as in this problem where the energy needed to heat 250 mL of soup can be calculated. Understanding how this property influences energy transfer is crucial when working with temperature changes and heat calculations in everyday scenarios.
Energy Transfer
Energy transfer is the process of energy moving from one place or object to another. In our example, this involves the transfer of electrical energy from the immersion heater to the water. Using the formula, \[ Q = mc\Delta T \]where \(Q\) is the heat energy transferred, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the temperature change, we can calculate how much energy is required to heat the water.
The transfer involves two main quantities:
  • The mass of the water (0.25 kg in this case, derived from 250 mL)
  • The specific heat capacity (4186 J/kg°C for water)
Multiplying these by the temperature change \(\Delta T = 60°C\), we calculate the energy to be 62,790 J. This calculation highlights how energy is systematically transferred and is useful to determine real-life applications such as cooking or heating.
Temperature Change
Understanding temperature change is vital when studying thermodynamics. It is simply the difference between the final and initial temperatures of a system. In the case of heating water from 15°C to 75°C, the temperature change, represented as \(\Delta T\), is 60°C.
When dealing with temperature change calculations, a few key points are considered:
  • Know the initial and final temperatures.
  • Calculate the difference to find \(\Delta T\).
This measurement is essential because it directly correlates with the amount of energy required to effect the change. Larger temperature changes usually require more energy, as demonstrated in the example. Essentially, by understanding the temperature change, you can predict and calculate the energy needed, leading to various practical applications in technology and daily life.

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Most popular questions from this chapter

(II) A \(58-\mathrm{kg}\) ice-skater moving at \(7.5 \mathrm{~m} / \mathrm{s}\) glides to a stop. Assuming the ice is at \(0^{\circ} \mathrm{C}\) and that \(50 \%\) of the heat generated by friction is absorbed by the ice, how much ice melts?

(II) Determine \((a)\) the work done and \((b)\) the change in internal energy of 1.00 \(\mathrm{kg}\) of water when it is all boiled to steam at \(100^{\circ} \mathrm{C}\) Assume a constant pressure of 1.00 atm.

(II) A 3.65-mol sample of an ideal diatomic gas expands adiabatically from a volume of \(0.1210 \mathrm{~m}^{3}\) to \(0.750 \mathrm{~m}^{3}\). Initially the pressure was 1.00 atm. Determine: \((a)\) the initial and final temperatures; (b) the change in internal energy; (c) the heat lost by the gas; \((d)\) the work done on the gas. (Assume no molecular vibration.)

(II) When a \(290-\mathrm{g}\) piece of iron at \(180^{\circ} \mathrm{C}\) is placed in a 95-g aluminum calorimeter cup containing \(250 \mathrm{~g}\) of glycerin at \(10^{\circ} \mathrm{C},\) the final temperature is observed to be \(38^{\circ} \mathrm{C}\) Estimate the specific heat of glycerin.

A soft-drink can contains about \(0.20 \mathrm{~kg}\) of liquid at \(5^{\circ} \mathrm{C}\). Drinking this liquid can actually consume some of the fat in the body, since energy is needed to warm the liquid to body temperature \(\left(37^{\circ} \mathrm{C}\right) .\) How many food Calories should the drink have so that it is in perfect balance with the heat needed to warm the liquid (essentially water)?

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