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Chapter 13: Problem 10

Water falls from a height of \(500 \mathrm{~m}\). What is the rise in temperature of water at the bottom if whole energy is used up in heating water ? (a) \(0.96^{\circ} \mathrm{C}\) (b) \(1.02^{\circ} \mathrm{C}\) (c) \(1.16^{\circ} \mathrm{C}\) (d) \(0.23^{*} \mathrm{C}\)

Short Answer

Expert verified
The rise in temperature is \(1.16^{\circ}\text{C}\) (option c).

Step by step solution

01

Understand the Concepts Involved

Water falling from a height has potential energy which is converted into heat energy when it hits the ground. The question asks us to find out how much the temperature of the water increases as a result of this energy conversion.
02

Identify Given Values and Constants

The height from which water falls is given as \(500 \mathrm{~m}\). The acceleration due to gravity \(g\) is \(9.8 \mathrm{~m/s^2}\). The specific heat capacity of water \(c\) is \(4.186 \mathrm{~J/g^ \circ C}\). Assume 1 gram of water for simplicity.
03

Calculate Potential Energy

Potential energy (PE) of the water is calculated using the formula:\[ \text{PE} = mgh \]where \(m\) is the mass of water (assumed as \(1\mathrm{~g}\)), \(g\) is the acceleration due to gravity, and \(h\) is the height. As \(m\) is \(1 \mathrm{~g}\), the potential energy is:\[ \text{PE} = 1 \times 9.8 \times 500 = 4900 \mathrm{~J} \]
04

Use Energy Conversion to Find Temperature Rise

The potential energy of \(4900 \mathrm{~J}\) gets converted into heat energy \(Q\) which increases the water's temperature. Use the specific heat capacity formula:\[ Q = mc\Delta T \]where \(\Delta T\) is the change in temperature. Rearrange to find \(\Delta T\):\[ \Delta T = \frac{Q}{mc} = \frac{4900 \mathrm{~J}}{1 \mathrm{~g} \times 4.186 \mathrm{~J/g^\circ C}} \approx 1.17^\circ \mathrm{C} \]This approximates to \(1.16^\circ \mathrm{C}\).
05

Select the Correct Option

Match the calculated temperature change with the given options. The closest match:Option (c) \(1.16^{\circ} \mathrm{C}\).

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Key Concepts

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

Potential Energy
When water is at a height, it possesses potential energy due to its position. This energy results from the gravitational pull of the Earth. The higher the object, the more potential energy it has.
For water falling from a height of 500 meters, the potential energy is given by the formula:
  • \[ \text{PE} = mgh \]
  • Where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.8 \( \mathrm{m/s^2} \)), and \( h \) is the height.
When the water falls and hits the ground, this stored energy is released.This concept of potential energy helps us understand how energy is converted from one form to another, a crucial principle in physics.
In this specific scenario, the potential energy of the water is converted into heat.
Specific Heat Capacity
The specific heat capacity is a measure of how much heat energy a substance can store per unit mass before its temperature changes.
For water, this value is \( 4.186 \mathrm{~J/g^\circ C} \). This means it takes 4.186 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.Understanding specific heat capacity is crucial for solving problems related to temperature changes, as it tells us how efficiently a substance can store heat energy.
  • A higher specific heat capacity means the substance can store more energy before its temperature rises.
  • For water, this makes it an excellent medium for storing and absorbing heat.
When energy, such as the potential energy from falling water, is converted into heat, this constant helps us determine how much the temperature of the water will increase.
Temperature Rise
The change in temperature of a substance occurs when it absorbs or releases heat energy.
In the case of falling water, the potential energy is converted entirely into heat energy upon impact with the ground, leading to a temperature rise.We can calculate this rise using the specific heat capacity formula:
  • \[ Q = mc\Delta T \]
Here:
  • \( Q \) is the heat energy (4900 J in this case),
  • \( m \) is the mass of the water (1 gram),
  • \( c \) is the specific heat capacity,
By rearranging the formula, \( \Delta T = \frac{Q}{mc} \), we find the increase in water temperature to be approximately 1.17掳C.
Thus, through energy conversion, the potential energy causes the water's temperature to rise, showing the direct relationship between energy and temperature change.

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Most popular questions from this chapter

A polished metal plate with a rough black spot on it is heated to about \(1400 \mathrm{~K}\) and quickly taken to a dark room. The spot will appear (a) darker than plate (b) brighter than plate (c) equally bright (d) equally dark

In heat transfer which method is based on gravitation (a) Natural convection (b) Conduction (c) Radiation (d) Stirrling of liquid

The rate of emission of radiation of a black body at temperature \(27^{\circ} \mathrm{C}\) is \(E_{1}\). If its temperature is increased to \(327^{\circ} \mathrm{C}\), the rate of emission of radiation is \(E_{2} .\) The relation between \(E_{1}\) and \(E_{2}\) is (a) \(E_{1}=24 E_{1}\) (b) \(E_{2}=16 E_{1}\) (c) \(E_{2}=8 E_{1}\) (d) \(E_{2}=4 E_{1}\)

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of \(0.1 \mathrm{~g}\) of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of the material of the second rod is \(1 / 4\) that of the first, the rate at which ice melts in \(\mathrm{gs}^{-1}\) will be (a) \(3.2\) (b) \(1.6\) (c) \(0.2\) (d) \(0.1\)

A steel scale measures the length of a copper wire as \(80.0 \mathrm{~cm}\), when both are at \(20^{\circ} \mathrm{C}\) (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at \(40^{\circ} \mathrm{C}\) ? (Given \(\alpha_{\text {nteel }}=11 \times 10^{-6}\) per \(^{\circ} \mathrm{C}\) and \(\alpha_{\text {copper }}=17 \times 10^{-6}\) per \(^{\circ} \mathrm{C}\) ) (a) \(80.0096 \mathrm{~cm}\) (b) \(80.0272 \mathrm{~cm}\) (c) \(1 \mathrm{~cm}\) (d) \(25.2 \mathrm{~cm}\)
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