CBSE Class 7 Science Advanced Physics Heat Long Answer Questions Set 01

Long Answer Questions

Question 1. The heat energy produced by the complete combustion of 25 g of a fuel is supplied to 500 g of water to increase the value of temperature from 35 °C to 55 °C. Find the calorific value of the fuel. (Specific heat capacity of water = 1 cal g⁻¹ °C⁻¹)
Answer:
Mass of water = 500 g, Δt = 55 − 35 = 20 °C
Heat absorbed by water Q = m s Δt = 500 × 1 × 20 = 10,000 cal
This is the heat produced by 25 g of fuel.
Calorific value = Q ÷ M = 10,000 ÷ 25 = 400 cal g⁻¹
In simple words: Find the heat taken by the water, then divide by the mass of fuel burnt.

Teacher's Note:
1. Equate heat from fuel to heat gained by water.
2. In SI: 400 × 4200 = 1.68 × 10⁶ J kg⁻¹.

Question 2. Describe an experiment to determine the melting point and boiling point of water.
Answer:
Place some crushed ice in a beaker and put a thermometer in it, held upright. Set the beaker on a stand and heat it with a burner.
The temperature rises to 0 °C. While the ice melts, the thermometer stays at 0 °C even though heat is supplied: melting point = 0 °C.
After all the ice has melted, the temperature of the water rises until it reaches 100 °C, where water starts to boil. The temperature stays at 100 °C until all the water has changed to steam: boiling point = 100 °C.
If temperature (Y-axis) is plotted against time of heating (X-axis), the graph has two flat portions, one at 0 °C and one at 100 °C.
During the flat parts the heat supplied is used only to change the state.
In simple words: Heat ice slowly with a thermometer in it and note the two temperatures that stay constant.

Teacher's Note:
1. Mention the graph and the two flat parts.
2. Draw the apparatus.

Question 3. Describe the construction of mercury thermometer.
Answer:
A very fine capillary tube of uniform bore, with a thin glass bulb at one end and a funnel at the other, is protected by a thick glass stem.
Mercury is poured through the funnel into the tube. Air bubbles that get trapped are driven out by warming the bulb in a water bath, and this is repeated until the bulb and tube are completely filled with mercury. The funnel end is cut and sealed.
To mark the scale, the bulb is placed in pure melting ice and the mercury level is marked as the lower fixed point (0 °C). The bulb is then held in steam and the level is marked as the upper fixed point (100 °C).
The distance between the two marks is divided into 100 equal parts, each part being 1 °C.
In simple words: Fill a fine tube with mercury, seal it, and mark 0 °C and 100 °C with ice and steam.

Teacher's Note:
1. Draw the labelled figure.
2. Mention the removal of air bubbles.

Question 4. Describe an experiment to show that water is a bad conductor of heat.
Answer:
Take a hard glass tube and put small pieces of ice, wrapped in copper wire gauze (so that they do not float), at the bottom.
Fill the tube up to 3/4 of its length with ice-cold water and clamp it in a slanting position.
Heat the tube near its mouth with a bunsen burner.
The water near the mouth starts boiling, but the ice at the bottom does not melt.
So heat is not conducted down through the water: water is a bad conductor of heat.
In simple words: The top water boils but the ice below stays unmelted.

Teacher's Note:
1. Heat at the top, not the bottom.
2. Draw the figure.

Question 5. Describe the construction of calorimeter.
Answer:
A calorimeter is a highly polished cylindrical copper vessel used to measure the quantity of heat.
The copper vessel is placed inside a wooden box, and the space between them is filled with a bad conductor of heat such as cork or sawdust.
The box has a lid of ebonite or another bad conductor with two holes: a stirrer passes through one and a thermometer through the other.
Principle: when a hot body is in contact with a colder body, the hot body loses heat and the cold body gains it.
Use: a known mass of water is taken in the calorimeter and its temperature noted. The hot substance is added, the mixture is stirred and the final temperature is read. Equating heat lost and heat gained gives the specific heat of the substance.
In simple words: A polished copper cup, insulated in a wooden box, with a stirrer and thermometer.

Teacher's Note:
1. Polished copper reduces radiation loss.
2. Cork reduces conduction loss.

Question 6. Describe an experiment to show that the solids expand on heating.
Answer:
Take Gravesand's ring and ball: the internal diameter of the ring equals the external diameter of the ball.
The cold ball just slips through the ring.
Heat the ball with a spirit lamp for some time and place it on the ring again; it does not pass through, showing that the ball has expanded.
Allow the ball to cool to room temperature and place it on the ring; it passes through easily, showing that solids contract on cooling.
In simple words: A heated ball becomes too big to pass through the ring.

Teacher's Note:
1. Describe the ball before and after heating.
2. Draw figures 3.11(a) and (b).

Question 7. Explain the anomalous expansion of water with Hope's apparatus.
Answer:
Water at 0 °C contracts on heating up to 4 °C, so water has minimum volume and maximum density at 4 °C. After 4 °C it expands. This unusual behaviour is the anomalous expansion of water.
Hope's apparatus: a tall cylinder of water has a trough of freezing mixture (common salt and ice 1 : 3) around its middle, and two thermometers, one near the top and one near the bottom. The cylinder is filled with water at 10 °C.
The water near the cooling mixture cools and its density increases, so it sinks to the bottom. When it reaches 4 °C it is densest and stays at the bottom, so the lower thermometer reads 4 °C and stays there.
The water in the middle cools below 4 °C, becomes lighter and rises. The upper thermometer reads 0 °C and a thin layer of ice forms on top.
Because of this, ponds freeze only at the surface and aquatic life survives in the 4 °C water below.
In simple words: Water is densest at 4 °C, so the cold water at 4 °C sinks and the freezing water floats on top.

Teacher's Note:
1. Draw Hope's apparatus.
2. Mention the fish in frozen ponds.

Question 8. (a) Write the relation between heat capacity and specific heat capacity.
(b) 7000 J of heat energy is supplied to a copper sphere of mass 90 g. What is the raise in temperature of the sphere? (Specific heat capacity of copper = 0.09 cal g⁻¹ °C⁻¹)
Answer:
(a) Heat capacity = mass × specific heat capacity, that is, C = m s.
(b) Heat Q = 7000 J = 7000 ÷ 4.2 = 1666.7 cal (taking 1 cal = 4.2 J).
m = 90 g, s = 0.09 cal g⁻¹ °C⁻¹, so heat capacity = 90 × 0.09 = 8.1 cal °C⁻¹
Rise in temperature Δt = Q ÷ (m s) = 1666.7 ÷ 8.1 = about 206 °C (205.8 °C).
In simple words: Convert joule to calorie, then divide the heat by the heat capacity of the sphere.

Teacher's Note:
1. Use the same unit for Q and s.
2. Δt = Q ÷ (m s).

Question 9. Is it possible to raise the temperature of water by heating it in a container made up of paper? [Temperature required to burn the paper is more than 100 °C]
Answer:
Yes, it is possible.
The paper container is in contact with the water, and the heat supplied to the paper is absorbed by the water. The water takes heat from the paper, so the paper stays at about the same temperature as the water (up to 100 °C).
Since this is less than the temperature needed to burn the paper, the paper does not burn while water is present.
In simple words: The water keeps taking the heat away, so the paper never gets hot enough to burn.

Teacher's Note:
1. Water boils at 100 °C and cannot become hotter.
2. The paper burns only when the water is gone.

Question 10. Rings made up of copper and iron are fixed as shown in the figure. On heating the rings, explain what happens to them, if the thickness of the rings are negligible.

Answer:
Both rings expand on heating, but copper expands more than iron.
The copper ring is inside the iron ring and tries to become larger than the space inside the iron ring, so it presses outward against the iron ring.
So the iron ring is stretched and breaks (cracks).
In simple words: The inner copper ring grows more than the outer iron ring and breaks it.

Teacher's Note:
1. Copper expands more than iron.
2. The figure shows copper inside iron.

Question 11. Explain why a clinical thermometer is exclusively used to measure the human body temperature and a laboratory thermometer is used to measure the temperatures of different substances but not to measure the human body temperature.
Answer:
The temperature of the human body normally stays between about 35 °C and 43 °C, so the clinical thermometer is made for this short range, with fine divisions, and it has a constriction that keeps the reading. It cannot measure other temperatures because its scale ends at 43 °C.
A laboratory thermometer reads from about −10 °C to 110 °C with each division 1 °C. Small variations in body temperature (a fraction of a degree) cannot be read accurately on it, and it has no constriction, so it is not used for the human body.
In simple words: A clinical thermometer has a short, fine scale; a laboratory thermometer has a wide, coarse scale.

Teacher's Note:
1. Mention range, accuracy and constriction.
2. Both are mercury thermometers.

Question 12. Two beakers A and B contain water at different temperatures. When 1 litre of water from beaker A is mixed with 2 litre of water from beaker B, the equilibrium temperature is 16 °C and when 2 litre of water from beaker A is mixed with 2 litre of water from beaker B, then the equilibrium temperature is 14 °C. Determine the temperature of water in the beakers A and B.
Answer:
Let the temperatures of A and B be t_A and t_B, with t_A less than t_B. Heat gained = heat lost, and water has the same specific heat, so the masses can be replaced by volumes.
First mixing: 1 × (16 − t_A) = 2 × (t_B − 16), so t_A + 2t_B = 48 ... (1)
Second mixing: 2 × (14 − t_A) = 2 × (t_B − 14), so t_A + t_B = 28 ... (2)
Subtracting (2) from (1): t_B = 20 °C, and then t_A = 28 − 20 = 8 °C.
So the temperature of beaker A is 8 °C and of beaker B is 20 °C.
Check: (1 × 8 + 2 × 20) ÷ 3 = 16 °C and (2 × 8 + 2 × 20) ÷ 4 = 14 °C.
In simple words: Write heat lost = heat gained for both mixings and solve the two equations.

Teacher's Note:
1. Subtract the two equations.
2. Always verify by mixing the values back.

Question 13. Two beakers contain hot water of the same temperature. One beaker is placed in another beaker containing cold water and the other beaker in front of a rotating fan. It is observed that the beaker placed infront the fan cools faster. Explain the reason.
Answer:
When the beaker stands in cold water, heat is lost only by conduction through the glass into the water, which warms up and slows down further cooling.
In front of a fan, air keeps moving past the beaker and carries the heat away by forced convection, and it also increases evaporation from the hot water surface, which takes away more heat.
So the heat transfer rate is higher and the beaker in front of the fan cools faster.
In simple words: Moving air carries heat away much faster than still cold water.

Teacher's Note:
1. Forced convection speeds up cooling.
2. Evaporation also helps.

Question 14. A test tube is filled with some water. A small piece of wax is fixed at the bottom of the test tube and a small piece of wax is made to float on the water. What happens to the wax at the top and bottom of the test tube if the test tube is heated at the top, middle and bottom?
Answer:
Heated at the top: the wax floating on top melts, but the wax at the bottom does not, because hot water stays at the top, there is no convection downward and water is a bad conductor.
Heated at the middle: the wax at the top melts because the hot water rises by convection, but the wax at the bottom stays solid.
Heated at the bottom: hot water rises by convection currents, so the wax at the bottom and the wax at the top both melt.
In simple words: Heat goes upward by convection, not downward, so heating at the bottom warms the whole tube.

Teacher's Note:
1. Water is a bad conductor.
2. Convection works upward only.

Question 15. Why is hot water more effective than cold water in extinguishing fire?
Answer:
Hot water needs less heat to change into steam than cold water does.
The steam formed above the fire pushes away the oxygen and cuts off the oxygen supply to the fire, so the fire goes out.
Hot water also turns into vapour more quickly, so it is more effective.
In simple words: Hot water turns into steam more quickly and the steam smothers the fire.

Teacher's Note:
1. Fire needs oxygen.
2. Steam blocks the oxygen.