Chapter 14 - Measurement and Effects of Heat
1. A. Whom should I pair with?
| A | B |
|---|---|
a. Temperature of a healthy human body | (i) 296 K |
b. Boiling point of water | (ii) 98.6⁰ F |
c. Room temperature | (iii) 0⁰ C |
d. Freezing point of water | (iv) 212⁰ F |
Ans:
| A | B |
|---|---|
a. Temperature of a healthy human body | (ii) 98.6⁰ F |
b. Boiling point of water | (iv) 212⁰ F |
c. Room temperature | (i) 296 K |
d. Freezing point of water | (iii) 0⁰ C |
B. Who is telling the truth?
a. The temperature of a substance is measured in Joules.
Ans: It is lying as the temperature of a substance is measured in ⁰C or Kelvin or ⁰F.
b. Heat flows from an object at higher temperature to an object at lower temperature.
Ans: It is telling the truth.
c. Joule is the unit of heat.
Ans: It is telling the truth.
d. Objects contract on heating.
Ans: It is lying as objects expand on heating.
e. Atoms of a solid are free.
Ans: It is lying as atoms of a solid are bound tightly to each other.
f. The average kinetic energy of atoms in a hot object is less than the average kinetic energy of atoms in a cold object.
Ans: It is lying as the average kinetic energy of atoms in a hot object is more than the average kinetic energy of atoms in a cold object.
C. You will find it if you search.
a. A thermometer is used to measure ……….
Ans: temperature
b. The apparatus used to measure heat is called a ………….
Ans: calorimeter
c. Temperature is the measure of the ……….. kinetic energy of the atoms in a substance.
Ans: average
d. The heat contained in a substance is the measure of the ………….. kinetic energy of atoms in the substance.
Ans: total
2. Nishigandha kept a vessel containing all the ingredients for making tea in a solar cooker. Shivani kept a similar vessel on a stove. Whose tea will be ready first and why?
Ans: Because the sun is far from the earth, the stove will provide more heat to the vessel for a given period of time than the sun. Hence, the tea made by Shivani using a stove will be ready first.
3. Write brief answers.
a. Describe a clinical thermometer. How does it differ from the thermometer used in the laboratory?
Ans:
(i) A clinical thermometer consists of a narrow glass tube having a bulb at one end.
(ii) The part of the tube along with the bulb is filled with mercury or alcohol. As mercury is harmful to health, alcohol is preferred.
(iii) The rest of the tube is evacuated and the other end of the tube is closed.
(iv) The bulb is kept in contact with the body of a person whose body temperature is to be measured. Thus, the temperature of the bulb becomes the same as that of the body of the person.
(v) If the body temperature is high, because of the increased temperature, the alcohol inside the tube expands and it rises in the tube. The body temperature can be obtained from the level of the alcohol inside the tube.
(vi) As the clinical thermometer is mainly used to measure body temperature, the range of this thermometer is marked from 35 °C to 42 °C..
(vii) The thermometer which is used in laboratories is constructed in a similar manner except that the thermometers in the laboratory can measure temperatures ranging from 40°C to 110°C or larger in some cases.
b. What is the difference between heat and temperature? What are their units?
Ans:
c. Explain the construction of a calorimeter. Draw the necessary figure.
Ans: A calorimeter is a device used to measure heat content of an object.
Construction:
(i) A calorimeter consists of two vessels similar to those of a thermos flask. The inner vessel is made of copper.
(ii) The space between the inner vessel and the outer vessel is filled with insulators like air. A heat-resistant ring connects both vessels so that no heat is transferred between the inner vessel and its surroundings.Thus, the inner vessel is thermally isolated from its surroundings.
(iii) Both the vessels have a single heat-resistant lid fitted on top. A thermometer and a stirrer are inserted into the inner vessel through two holes in the lid.
(iv) The thermometer is used to record the temperature of the liquid, and the stirrer is used for stirring the liquid.
d. Explain why rails have gaps at specific distances.
Ans:
(i) During summer, due to the change in temperature, the length of the rails increases.
(ii) If the rails are made without gaps in between, then after expansion, the rails may get distorted in the summer, leading to rail accidents.
(iii) Hence, to accommodate this change in length, the rails have gaps at specific distances.
e. Explain with the help of formulae the expansion coefficients of liquid and gas.
Ans: Coefficient of expansion for liquids:
(i) Liquids do not have definite shape or size, but they have definite volume. Hence, liquids possess volumetric expansion coefficient.
(ii) Let liquid of initial volume V, expands to a volume V, when heated. If ΔT is the change in temperature, then, V2 = V₁ (1 + βΔT). Here, β is volumetric expansion coefficient of the liquid.
Coefficient of expansion for gases:
(i) Gases do not possess definite shape, size or volume. They occupy the volume of the container.
(ii) Gases expand on heating when kept in a open container. If the gas is kept in a closed container and heated, then the pressure on the gas increases while the volume remains unchanged.
(iii) Thus, the expansion of a gas is measured under constant pressure and the expansion coefficient is termed as constant pressure expansion coefficient.
(iv) If a gas is heated in a beaker with a movable piston attached to the beaker, then pressure remains constant while volume increases. In such a case, volumetric expansion is given as,
V₂ = V₁ (1 + βΔT)
Here, V₂ = Final volume of the gas,
V₁ = Initial volume of the gas,
β = Constant pressure expansion coefficient,
ΔT = Change in the temperature.
4. Solve the following examples.
a. What must be the temperature in Fahrenheit so that it will be twice its value in Celsius?
Given:
Temperature in Fahrenheit (F) = 2 x temperature in Celsius (C)
i.e. F = 2C
∴ C = F/2
To find:
Temperature in Fahrenheit (F)
Formula:
$$ {\frac{F\;-\;32}9=\frac C{5}\\} $$
Solution:
Substituting for C in formula,
$$ {\frac{F\;-\;32}9=\frac F{2\times5}\\\frac{F\;-\;32}9=\frac F{10}\\10(F\;-\;32)\;=\;9F\\10F\;-\;320\;=\;9F\\10F\;-\;9F\;=\;320\\F\;=\;320^\circ F} $$
Ans: The temperature 320°F is twice its value in Celsius.
b. A bridge is made from 20 m long iron rods. At temperature 18⁰ C, the distance between two rods is 0.4 cm. Up to what temperature will the bridge be in good shape?
(Note: The bridge would be in good shape until the expansion in the rods does not exceed the gap between them.)
Given:
Initial length of rod (l₁) = 20 m
Initial temperature (T₁) = 18° C
Final length of rod (l₁) = 20m + 0.4cm
= (20 + 0.004) m
= 20.004m
Coefficient of linear expansion for iron (iron) = 11.5 x 10⁻⁶ (1/°C)
To find:
Final temperature (T₂)
Formula:
l₂ = l₁ (1 + λΔT)
Solution:
From formula,
$$ {20.004\;=\;20\;\lbrack1\;+\;(11.5\;\times\;10^{-6})\triangle T\rbrack\\؞\;\frac{20.004}{20}=1\;+\;(11.5\;\times\;10^{-6})\triangle T\\؞\;(11.5\;\times\;10^{-6})\triangle T\;=\;\frac{20.004}{20}\;-\;1\\؞\;(11.5\;\times\;10^{-6})\triangle T\;\;=\;1.0002\;-\;1\\؞\;(11.5\;\times\;10^{-6})\triangle T\;\;=\;0.0002\\؞\;\triangle T\;=\;\frac{0.0002}{11.5\;\times\;10^{-6}}\\؞\;\triangle T\;=\;\frac{200\;\times\;10^{-6}}{11.5\;\times\;10^{-6}}\\؞\;\triangle T\;=\;\frac{200}{11.5}\\؞\;\triangle T\;=\;17.4\\\\T_2\;-\;T_1\;=\;17.4\\؞\;T_2\;=\;17.4\;+\;T_1\\؞\;T_2\;=\;17.4\;+\;18\\؞\;T_2\;=35.4^\circ C\\\\\\\\\\\\\\} $$
Ans: The bridge will be in good shape upto the temperature of 35.4 °C.
c. At 15⁰C the height of Eiffel tower is 324 m. If it is made of iron, what will be the increase in length in cm, at 30⁰C?
Given:
Initial height of Eiffel tower (l₁) = 324 m,
initial temperature (T₁) = 15° C
final temperature (T₂) = 30 °C
coefficient of linear expansion for iron (iron) = 11.5 x 10⁻⁶ (1/°C)
To find:
Increase in the height (l₂ – l₁)
Formula:
l₂ = l₁ (1 + λΔT)
Solution:
From formula,
l₂ = l₁ (1 + λΔT)
l₂ – l₁ = (1 + λΔT)
∴ l₂ – l₁ = 11.5 x 10⁻⁶ x 324 (30 – 15)
∴ l₂ – l₁ = 11.5 x 10⁻⁶ × 324 (15)
∴ l₂ – l₁ = 11.5 x 10⁻⁶ × 4860
∴ l₂ – l₁ = 55890 × 10⁻⁶
∴ l₂ – l₁ = 0.056m
∴ l₂ – l₁ = (0.056 × 100) cm
∴ l₂ – l₁ = 5.6cm
Ans: Increase in the height of Eiffel tower will be 5.6 cm.
d. Two substances A and B have specific heats c and 2 c respectively. If A and B are given Q and 4Q amounts of heat respectively, the change in their temperatures is the same. If the mass of A is m, what is the mass of B?
Given:
Specific heat of substance A (cA) = c
Specific heat of substance B (cB) = 2c
Amount of heat given to A (QA) = Q
Amount of heat given to B (QB) = 4Q
ΔTA = ΔTB
mass of substance A (mA) = m
To find:
Mass of substance B (mB)
Formula:
Q = m × c × ΔT
Solution:
From formula,
$$ {\Delta T\;=\;\frac Q{m\;\times\;c}\\} $$
$$ {\Delta T\;=\;\frac Q{m\;\times\;c}\\As\;\Delta T_{A\;}\;=\;\Delta T_B\;\;\;…(Given)\\؞\;\;\frac{Q_A}{m_A\;\times\;c_A}\;=\;\frac{Q_B}{m_B\;\times\;c_B}\\؞\;\frac Q{m\;\times\;c}\;=\;\frac{4Q}{m_B\;\times\;2c}\\؞\;\frac1m\;=\;\frac2{m_B}\\؞\;m_B\;=\;2m\\} $$
Ans: The mass of substance B is 2m.
e. When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by 10⁰C. What is the specific heat of the substance?
Given:
Mass of the substance (m) = 3 kg = 3 × 10³g
amount of heat given (Q) = 600 cal
change in temperature (ΔT) = 10⁰ C
To find:
Specific heat of substance (c)
Formula:
Q = m × c × ΔT
Solution:
From formula,
$$ {600\;=\;3\;\times\;10⁻³\;\times\;c\;\times\;10\;\\c\;=\;\frac{600}{3\;\times\;10^4}\\؞\;c\;=\;\frac1{50}\\؞\;c\;=\;0.02\;cal/g⁰C\\\\\\\\\\\\\\} $$
Ans: The specific heat of the substance is 0.02 cal/g°C.