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Thermal Properties of Matter - Specific Heat Capacity and Calorimetry

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Heat (QQ) is the form of energy that is transferred between systems or objects with different temperatures. It is measured in Joules (JJ) or Calories (calcal).

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Heat Capacity (SS) of a substance is the amount of heat required to raise its temperature by 1∘C1^{\circ}C or 1 K1 \text{ K}.

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Specific Heat Capacity (cc) is defined as the amount of heat per unit mass required to raise the temperature by 1∘C1^{\circ}C. It depends on the nature of the substance and its temperature.

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Molar Specific Heat Capacity (CC) is the heat required to raise the temperature of 1 mole1 \text{ mole} of a substance by 1 K1 \text{ K}.

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Principle of Calorimetry: In an isolated system, the heat lost by the hot body is equal to the heat gained by the cold body, i.e., Heat Lost=Heat Gained\text{Heat Lost} = \text{Heat Gained}.

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Latent Heat (LL) is the heat required to change the state of a unit mass of a substance without any change in its temperature. For fusion, it is LfL_f, and for vaporization, it is LvL_v.

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Water Equivalent (ww) of a body is the mass of water that would absorb or release the same amount of heat as the body for the same rise or fall in temperature.

📐Formulae

Q=mcΔTQ = mc\Delta T

S=ΔQΔT=mcS = \frac{\Delta Q}{\Delta T} = mc

C=1μΔQΔTC = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}

Q=mLQ = mL

m1c1(T1−T)=m2c2(T−T2)m_1 c_1 (T_1 - T) = m_2 c_2 (T - T_2)

💡Examples

Problem 1:

A sphere of aluminum of mass 0.047 kg0.047 \text{ kg} is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100∘C100^{\circ}C. It is then immediately transferred to 0.14 kg0.14 \text{ kg} copper calorimeter containing 0.25 kg0.25 \text{ kg} of water at 20∘C20^{\circ}C. The lowest temperature attained by water is 23∘C23^{\circ}C. Calculate the specific heat capacity of aluminum. (Given: Specific heat of copper ccu=0.386×103 J kg−1K−1c_{cu} = 0.386 \times 10^3 \text{ J kg}^{-1} \text{K}^{-1}, Specific heat of water cw=4.18×103 J kg−1K−1c_w = 4.18 \times 10^3 \text{ J kg}^{-1} \text{K}^{-1})

Solution:

Let calc_{al} be the specific heat of aluminum. Heat lost by aluminum sphere: Q1=malcal(Tinitial−Tfinal)=0.047×cal×(100−23)Q_1 = m_{al} c_{al} (T_{initial} - T_{final}) = 0.047 \times c_{al} \times (100 - 23). Heat gained by water and calorimeter: Q2=(mwcw+mcuccu)(Tfinal−Tinitial,water)=(0.25×4.18×103+0.14×0.386×103)(23−20)Q_2 = (m_w c_w + m_{cu} c_{cu})(T_{final} - T_{initial, water}) = (0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3)(23 - 20). Equating Q1=Q2Q_1 = Q_2: 0.047×cal×77=(1045+54.04)×30.047 \times c_{al} \times 77 = (1045 + 54.04) \times 3. Resulting in cal≈911 J kg−1K−1c_{al} \approx 911 \text{ J kg}^{-1} \text{K}^{-1}.

Explanation:

We apply the principle of calorimetry where the heat lost by the hot aluminum sphere is equated to the heat gained by the water and the copper container (calorimeter) until they reach thermal equilibrium at 23∘C23^{\circ}C.

Problem 2:

How much heat is required to convert 10 g10 \text{ g} of ice at 0∘C0^{\circ}C to steam at 100∘C100^{\circ}C? (Given: Lf=80 cal/gL_f = 80 \text{ cal/g}, Lv=540 cal/gL_v = 540 \text{ cal/g}, cw=1 cal/g∘Cc_w = 1 \text{ cal/g}^{\circ}C)

Solution:

Total Heat Q=Qmelt+Qheat+QvapQ = Q_{melt} + Q_{heat} + Q_{vap}. Q=mLf+mcwΔT+mLvQ = mL_f + mc_w\Delta T + mL_v. Q=(10×80)+(10×1×100)+(10×540)=800+1000+5400=7200 calQ = (10 \times 80) + (10 \times 1 \times 100) + (10 \times 540) = 800 + 1000 + 5400 = 7200 \text{ cal}.

Explanation:

The process involves three stages: melting the ice at constant temperature (mLfmL_f), raising the water temperature from 0∘C0^{\circ}C to 100∘C100^{\circ}C (mcΔTmc\Delta T), and then vaporizing the water at constant temperature (mLvmL_v).

Specific Heat Capacity and Calorimetry Revision - Class 11 Physics CBSE