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Question 13 Marks
(a) What is a calorimeter?
(b) Name the material of which a calorimeter is generally made.
(c) A calorimeter of mass 25.5 g has specific heat capacity of $0.42 Jg ^{-10} C ^{-1}$. Calculate its heat capacity.
Answer
(a) A small metal vessel in the form of beaker, used in calorimetric experiments is called calorimeter.
(b) Calorimeter is generally made from copper metal.
(c) Heat capacity of calorimeter $=$ Mass $\times$ Specific heat capacity.
$=25.5 g \times 0.42 Jg ^{-10} C ^{-1}=10.71 J^{\circ} C ^{-1}$.
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Question 23 Marks
The melting point of naphthalene is $80^{\circ} C$ and the room temperature is $30^{\circ} C$. A sample of liquid naphthalene at $100^{\circ} C$ is cooled down to the room temperature. Draw a temperature time graph to represent this cooling. In the graph, mark the region which corresponds to the freezing process.
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Question 33 Marks
0.04 kg of water at $80^{\circ} C$ is poured in a copper calorimeter with a stirrer of mass 0.32 kg and initially at $25{ }^{\circ} C$. After stirring, the final temperature recorded is $55^{\circ} C$. If the specific heat capacity of water is $4200 Jkg ^{-1}{ }^{\circ} C ^{-1}$, answer the following questions :
(a) What is the quantity of heat released when the temperature of 1 kg of water falls through $1^{\circ} C$ ?
(b) Calculate the total heat energy released by water in the above question on cooling from $80^{\circ} C$ to $55{ }^{\circ} C$.
(c) If all the heat energy released by water is absorbed by calorimeter, calculate the specific beat capacity of calorimeter,
Answer
(a) 1 kg of water on cooling through $1^{\circ} C$ will release 4200 J of heat energy.
(b) Total heat energy released by water $=m c \theta_{ F }=0.04 kg \times 4200 Jkg ^{-1}{ }^{\circ} C ^{-1} \times(80-55){ }^{\circ} C =4200 J$.
(c) Heat energy absorbed by calorimeter $=m c 0_{ R }=0.32 kg \times c \times(55-25)^{\circ} C =9.6 c kg { }^{\circ} C$
Heat absorbed by calorimeter $=$ Heat released by water
$\Rightarrow 9.6 c kg{ }^{\circ} C=4200 J$
Specific heat capacity of calorimeter $( c )=\frac{4200}{9.6} \frac{J}{ kg ^{\circ} C }=437.5 J kg ^{-1}{ }^{\circ} C ^{-1}$.
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Question 43 Marks
40 g of water at $60^{\circ} C$ is poured into a vessel containing 50 g of water at $20^{\circ} C$. The final temperature recorded is $30^{\circ} C$. Calculate the thermal capacity of the vessel. (Take specific heat capacity of water as $4.2 J g ^{-10} C ^{-1}$ )
Answer
Mass of hot water $=40 g$; Temperature of hot water $=60^{\circ} C$
Mass of cold water $=50 g$, Temperature of cold water $=20^{\circ} C$;
Final temperature $=30^{\circ} C$
Thermal capacity of vessel $(x)=$ ?
Specific heat capacity of water $=4.2 Jg ^{1{ }^{\circ}} C ^{-1}$
$\therefore$ Rise in temperature of cold water and vessel $\left(\theta_{ R }\right)=(30-20)=10^{\circ} C$
Fall in temperature of hot water $\left(\theta_f\right)-(60-30)=30^{\circ} C$
$\therefore$ Heat lost by hot water $=m c \theta_{ f }=40 \times 4.2 \times 30=5040 J$
Heat gained by cold water $=m c \theta_{ R }=50 \times 4.2 \times 10=2100 J$
Heat guined by vessel $=m c \theta_{ R }=x \times 10$
Heat gained by vessel + Heat gained by cold water $=$ Heat lost by hot water
$\Rightarrow 10 x+2100=5040$
$\therefore x=\frac{5040-2100}{10}=294 J^{\circ} C ^{-1}$
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Question 53 Marks
400 g of wax at $10^{\circ} C$ is beated to $80^{\circ} C$, when it starts melting. On complete melting wax is further heated so that temperature rises to $130^{\circ} C$. Calculate (a) Heat energy required to bring the wax to its melting point (b) Heat energy required to melt the wax (c) Heat energy required to bring the molten wax to $130^{\circ} C$.
[sp, beat capacity of solid wax $=1.5 Jg ^{-10} C ^{-1}$; sp. heat capacity of liquid wax $=1.8 Jg ^{-19} C ^{-1}$ and sp . latent heat of wax $\left.=20 Jg ^{-1}\right]$
Answer
1. (a) Heat energy required to raise temperature of solid wax from $10^{\circ} C$ to $80^{\circ} C =m c \theta_{ R }$
$=400 g \times 1.5 Jg^{-1}{ }^{\circ} C^{-1} \times 70^{\circ} C=42000 J$
(b) Heat energy required to melt the wax at $80^{\circ} C$
$=mL=400 g \times 20 Jg^{-1}=8000 J$
(c) Heat energy required to raise temperature of molten wax from $80^{\circ} C$ to $130^{\circ} C$
$=m c \theta_{R}=400 g \times 1.8 Jg^{-10} C^{-1} \times 50^{\circ} C=36000 J$
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Question 63 Marks
(a) Why is melting ice a better coolant than water at zero degree celsius?
(b) Why does ice-cream feel more colder than water at $0^{\circ} C$ ?
(c) Why does the weather become warm, when it snows?
Answer
(a) It is because ice absorbs 336000 J heat energy per kg from the surroundings while melting, as compared to 1 kg of water $0^{\circ} C$. Hence, cooling produced by ice is much more than water at $0^{\circ} C$.
(b) Ice extracts $336 J / g$ more heat from the mouth compared to 1 gm of water at $0^{\circ} C$ on account of its highest sp. latent heat of Fusion. Thus, ice-cream feels colder than water at $0^{\circ} C$.
(c) As water vapour freezes, it releases large amount of energy into the atmosphere owing to its highest specific latent heat. Thus the weather becomes warm.
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Question 73 Marks
Explain the following:
(a) Water is sprinkled on the roads in the evening during hot summer.
(b) Water tubs are kept in warchouses storing fruits and vegetables in cold countries during winter.
(c) A wise farmer always waters his fields in the evening, if there is a forecast for frost.
Answer
(a) The water sprinkled on the ground absorbs large amount of heat from the ground and evaporates. This evaporation of water causes cooling of the ground.
(b) Water does not cool quickly due to its high specific heat capacity. It releases a large amount heat for a small drop in temperature. Otherwise, fruits and vegetables would become useless due to freezing of water molecules inside. Thus, the fruits and vegetables remain fresh during winter.
(c) When temperature falls below $0^{\circ} C$, water in the capillaries of plants would freeze and thus capillaries burst due to anomalous expansion of water on cooling. As a result plants would die and the crop would get destroyed. In order to save crops on such cold nights, farmers water their fields in the evening. As water has high specific heat capacity it liberates lot of heat energy as it cools and does not allow the temperature in the surrounding of plants to fall below $0^{\circ} C$.
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[3 Mark Question Answer] - Physics STD 10 Questions - Vidyadip