Vidaara.orgClass 11 · Physics
CodeVID-P11-10-CCS-01
Calorimetry & Change of State — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The specific heat of water is approximately:
- A.$900\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
- B.$4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
- C.$2100\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
- D.$334\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
2.
One calorie is equal to:
- A.$1\ \text{J}$
- B.$4.186\ \text{J}$
- C.$0.24\ \text{J}$
- D.$1000\ \text{J}$
3.
Latent heat of vaporisation refers to a change from:
- A.solid to liquid
- B.liquid to gas
- C.gas to solid
- D.solid to gas
4.
During a change of state, the heat supplied changes the:
- A.temperature
- B.molecular potential energy / bonds
- C.mass
- D.pressure only
5.
If the same heat is given to equal masses of two substances, the one with smaller specific heat will have a:
- A.smaller rise in temperature
- B.larger rise in temperature
- C.no change in temperature
- D.decrease in temperature
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define specific heat capacity and give its SI unit.
7.
Why does water act as a good coolant?
8.
Find the heat to melt 250 g of ice at $0\,^\circ$C. ($L_f=3.34\times10^5\ \text{J}\,\text{kg}^{-1}$)
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
State the principle of calorimetry and explain how it is used to find an unknown specific heat.
10.
How much heat is needed to convert 50 g of ice at $0\,^\circ$C to water at $30\,^\circ$C? ($L_f=3.34\times10^5$, $c_w=4186$)
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Draw and explain the temperature versus heat graph when ice at $-10\,^\circ$C is heated steadily until it becomes steam. Identify the plateaus and the role of latent heat.
Answer Key
Section A — Multiple Choice Questions
- (B) $4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
- (B) $4.186\ \text{J}$
- (B) liquid to gas
- (B) molecular potential energy / bonds
- (B) larger rise in temperature
Section B — Short Answer (2 marks)
- Heat needed to raise the temperature of unit mass by one degree; unit $\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$.
- Its high specific heat lets it absorb a large amount of heat for a small temperature rise.
- $Q=mL_f=0.25\times3.34\times10^5=8.35\times10^4\ \text{J}$.
Section C — Short Answer (3 marks)
- In an insulated system, heat lost by the hot body equals heat gained by the cold body. Equating $m_1c_1\Delta T_1=m_2c_2\Delta T_2$ lets the unknown $c$ be found from measured masses and temperatures.
- $Q=0.05\times3.34\times10^5+0.05\times4186\times30=1.67\times10^4+6.28\times10^3=2.30\times10^4\ \text{J}$.
Section D — Long Answer (5 marks)
- The graph rises (ice warming) to $0\,^\circ$C, stays flat during melting (latent heat of fusion absorbed at constant T), rises again (water warming) to $100\,^\circ$C, stays flat (longer) during boiling (latent heat of vaporisation), then rises (steam heating). The two horizontal plateaus mark changes of state where heat goes into breaking molecular bonds, not raising temperature.
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