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Thermal Properties of Matter - Temperature and Heat

Grade 11CBSEPhysics

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

🔑Concepts

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Temperature is a macroscopic property of a system which determines the direction of heat flow when two bodies are in thermal contact. Heat flows from a body at higher temperature to a body at lower temperature.

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A thermometer is a device used to measure temperature, based on properties like volume expansion of liquids or pressure changes in gases. Common scales include Celsius (∘C^{\circ}C), Fahrenheit (∘F^{\circ}F), and Kelvin (KK).

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Thermal Expansion: Most substances expand when heated. The increase in length is Linear Expansion (αα), in area is Areal Expansion (ββ), and in volume is Volume Expansion (γγ). The relationship between them is α=β2=γ3\alpha = \frac{\beta}{2} = \frac{\gamma}{3}.

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Specific Heat Capacity (ss or cc) is the amount of heat required to raise the temperature of a unit mass of a substance by 1∘C1^{\circ}C or 1K1 K. Q=msΔTQ = ms\Delta T.

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Calorimetry is the measurement of heat. The principle of calorimetry states that in an isolated system, the heat lost by the hot body is equal to the heat gained by the cold body: Heat Lost=Heat GainedHeat \, Lost = Heat \, Gained.

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Latent Heat (LL) is the heat absorbed or released by a substance during a change of phase at constant temperature. Q=mLQ = mL. Two types exist: Latent heat of fusion (LfL_f) and Latent heat of vaporization (LvL_v).

📐Formulae

tC100=tF−32180=T−273.15100\frac{t_C}{100} = \frac{t_F - 32}{180} = \frac{T - 273.15}{100}

Δl=αlΔT\Delta l = \alpha l \Delta T

ΔV=γVΔT\Delta V = \gamma V \Delta T

Q=msΔTQ = m s \Delta T

Q=mLQ = m L

γ=3α\gamma = 3\alpha

💡Examples

Problem 1:

Calculate the amount of heat required to convert 0.5 kg0.5 \, kg of ice at 0∘C0^{\circ}C to water at 20∘C20^{\circ}C. (Given: Latent heat of fusion of ice Lf=3.34×105 J/kgL_f = 3.34 \times 10^5 \, J/kg and specific heat of water sw=4186 J/kg⋅Ks_w = 4186 \, J/kg\cdot K).

Solution:

Total heat required Q=Q1+Q2Q = Q_1 + Q_2 where Q1Q_1 is for phase change and Q2Q_2 is for temperature rise. Q1=mLf=0.5×3.34×105=1,67,000 JQ_1 = m L_f = 0.5 \times 3.34 \times 10^5 = 1,67,000 \, J Q2=mswΔT=0.5×4186×(20−0)=41,860 JQ_2 = m s_w \Delta T = 0.5 \times 4186 \times (20 - 0) = 41,860 \, J Total Heat=167000+41860=208860 JTotal \, Heat = 167000 + 41860 = 208860 \, J

Explanation:

First, the ice melts at a constant temperature (0∘C0^{\circ}C) using latent heat. Then, the resulting water is heated from 0∘C0^{\circ}C to 20∘C20^{\circ}C using its specific heat capacity.

Problem 2:

An iron rod of length 2 m2 \, m is heated from 30∘C30^{\circ}C to 80∘C80^{\circ}C. Find the change in its length if the coefficient of linear expansion α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5} \, K^{-1}.

Solution:

Given: l=2 ml = 2 \, m, α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5} \, K^{-1}. First, find the temperature difference: 80−3050\begin{array}{r} 80 \\ - 30 \\ \hline 50 \end{array} So, ΔT=50 K\Delta T = 50 \, K. Δl=αlΔT=(1.2×10−5)×2×50\Delta l = \alpha l \Delta T = (1.2 \times 10^{-5}) \times 2 \times 50 Δl=1.2×10−5×100=1.2×10−3 m=1.2 mm\Delta l = 1.2 \times 10^{-5} \times 100 = 1.2 \times 10^{-3} \, m = 1.2 \, mm

Explanation:

The change in length is calculated using the formula for linear thermal expansion, where the change is proportional to the original length and the rise in temperature.