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Thermodynamics - Calorimetry

Grade 11ICSEPhysics

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

🔑Concepts

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Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature. Its S.I. unit is the Joule (JJ), and another common unit is the calorie (calcal), where 1 cal=4.186 J1 \text{ cal} = 4.186 \text{ J}.

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Specific Heat Capacity (cc) is the amount of heat energy required to raise the temperature of a unit mass of a substance by 1∘C1^\circ\text{C} or 1 K1 \text{ K}. For water, cw≈4200 J kg−1 K−1c_w \approx 4200 \text{ J kg}^{-1} \text{ K}^{-1} or 1 cal g−1 ∘C−11 \text{ cal g}^{-1} \text{ }^\circ\text{C}^{-1}.

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Thermal Capacity or Heat Capacity (C′C') is the heat required to raise the temperature of the entire body by 1∘C1^\circ\text{C}. It is calculated as C′=mcC' = mc.

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The Principle of Method of Mixtures (Calorimetry) states that for an isolated system with no heat loss to the surroundings: Heat Lost by Hot Body=Heat Gained by Cold Body\text{Heat Lost by Hot Body} = \text{Heat Gained by Cold Body}.

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Latent Heat (LL) is the heat energy absorbed or released during a change of phase at a constant temperature. Specific Latent Heat of Fusion for ice is Lf≈336 J g−1L_f \approx 336 \text{ J g}^{-1} (80 cal g−180 \text{ cal g}^{-1}), and Specific Latent Heat of Vaporization for steam is Lv≈2260 J g−1L_v \approx 2260 \text{ J g}^{-1} (540 cal g−1540 \text{ cal g}^{-1}).

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Water Equivalent (WW) of a body is the mass of water which absorbs or radiates the same amount of heat as the body for the same rise or fall in temperature: W=mccwW = \frac{mc}{c_w}.

📐Formulae

Q=mcΔTQ = mc\Delta T

C′=mc=QΔTC' = mc = \frac{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)

W=mccwW = \frac{mc}{c_w}

Power(P)=Qt=mcΔTt\text{Power} (P) = \frac{Q}{t} = \frac{mc\Delta T}{t}

💡Examples

Problem 1:

Calculate the amount of heat energy required to convert 5 g5 \text{ g} of ice at 0∘C0^\circ\text{C} into water at 20∘C20^\circ\text{C}. Given: Lice=336 J g−1L_{ice} = 336 \text{ J g}^{-1} and cwater=4.2 J g−1 K−1c_{water} = 4.2 \text{ J g}^{-1} \text{ K}^{-1}.

Solution:

Q1=mL=5×336=1680 JQ_1 = mL = 5 \times 336 = 1680 \text{ J} (Heat to melt ice), Q2=mcΔT=5×4.2×(20−0)=420 JQ_2 = mc\Delta T = 5 \times 4.2 \times (20 - 0) = 420 \text{ J} (Heat to raise water temperature). Total Q=Q1+Q2=1680+420=2100 JQ = Q_1 + Q_2 = 1680 + 420 = 2100 \text{ J}.

Explanation:

The total heat is the sum of the heat required for the phase change (latent heat) and the heat required for the temperature rise (sensible heat).

Problem 2:

A copper calorimeter of mass 50 g50 \text{ g} contains 100 g100 \text{ g} of water at 30∘C30^\circ\text{C}. If 20 g20 \text{ g} of iron nails at 100∘C100^\circ\text{C} are dropped into it, find the final temperature. Assume ccopper=0.4 J g−1 K−1c_{copper} = 0.4 \text{ J g}^{-1} \text{ K}^{-1}, ciron=0.47 J g−1 K−1c_{iron} = 0.47 \text{ J g}^{-1} \text{ K}^{-1}, and cwater=4.2 J g−1 K−1c_{water} = 4.2 \text{ J g}^{-1} \text{ K}^{-1}.

Solution:

Let final temperature be TT. Heat lost by iron: Qlost=20×0.47×(100−T)Q_{lost} = 20 \times 0.47 \times (100 - T). Heat gained by calorimeter and water: Qgain=(50×0.4+100×4.2)×(T−30)Q_{gain} = (50 \times 0.4 + 100 \times 4.2) \times (T - 30). Equating Qlost=QgainQ_{lost} = Q_{gain}: 9.4(100−T)=(20+420)(T−30)⇒940−9.4T=440T−13200⇒449.4T=14140⇒T≈31.46∘C9.4(100 - T) = (20 + 420)(T - 30) \Rightarrow 940 - 9.4T = 440T - 13200 \Rightarrow 449.4T = 14140 \Rightarrow T \approx 31.46^\circ\text{C}.

Explanation:

Applying the Principle of Calorimetry, the heat energy lost by the hot iron nails is absorbed by both the copper calorimeter vessel and the water inside it until thermal equilibrium is reached.

Calorimetry - Revision Notes & Key Formulas | ICSE Class 11 Physics