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Physics - Calorimetry

Grade 10ICSE

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

🔑Concepts

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Heat is the internal energy of molecules in motion, measured in Joules (JJ) or Calories (calcal). 1 cal=4.186 J≈4.2 J1 \text{ cal} = 4.186 \text{ J} \approx 4.2 \text{ J}.

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Temperature is the degree of hotness or coldness of a body, representing the average kinetic energy of its molecules.

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Heat Capacity (C′C') is the amount of heat energy required to raise the temperature of the entire body by 1∘C1^\circ\text{C} or 1 K1 \text{ K}. Unit: J K−1J \text{ K}^{-1} or J ∘C−1J \text{ } ^\circ\text{C}^{-1}.

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Specific Heat Capacity (cc) is the heat required to raise the temperature of a unit mass (1 kg1 \text{ kg} or 1 g1 \text{ g}) of a substance by 1 K1 \text{ K} or 1∘C1^\circ\text{C}. Unit: J kg−1 K−1J \text{ kg}^{-1} \text{ K}^{-1}.

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Water has an unusually high specific heat capacity (cwater=4200 J kg−1 K−1c_{water} = 4200 \text{ J kg}^{-1} \text{ K}^{-1}), making it an excellent coolant.

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Principle of Calorimetry: For an insulated system, Heat energy lost by the hot body = Heat energy gained by the cold body.

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Latent Heat is the heat energy absorbed or released during a change of phase (solid to liquid or liquid to gas) without any change in temperature.

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Specific Latent Heat of Fusion (LfL_f) of ice is 336 J g−1336 \text{ J g}^{-1} or 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1}.

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Specific Latent Heat of Vaporization (LvL_v) of steam is 2260 J g−12260 \text{ J g}^{-1} or 2.26×106 J kg−12.26 \times 10^6 \text{ J kg}^{-1}.

📐Formulae

Q=mcΔTQ = mc\Delta T

C′=QΔT=mcC' = \frac{Q}{\Delta T} = mc

Q=mLQ = mL

m1c1(T1−T)=m2c2(T−T2)m_1c_1(T_1 - T) = m_2c_2(T - T_2)

Heat lost by hot body=Heat gained by cold body\text{Heat lost by hot body} = \text{Heat gained by cold body}

💡Examples

Problem 1:

Calculate the amount of heat energy required to raise the temperature of 200 g200 \text{ g} of copper from 20∘C20^\circ\text{C} to 70∘C70^\circ\text{C}. (Specific heat capacity of copper = 0.39 J g−1 K−10.39 \text{ J g}^{-1} \text{ K}^{-1})

Solution:

Given: m=200 gm = 200 \text{ g}, c=0.39 J g−1 K−1c = 0.39 \text{ J g}^{-1} \text{ K}^{-1}, ΔT=70−20=50∘C\Delta T = 70 - 20 = 50^\circ\text{C}. Using the formula Q=mcΔTQ = mc\Delta T: Q=200×0.39×50=3900 JQ = 200 \times 0.39 \times 50 = 3900 \text{ J}.

Explanation:

The formula Q=mcΔTQ = mc\Delta T is used because there is a change in temperature without a phase change.

Problem 2:

How much heat is required to convert 10 g10 \text{ g} of ice at 0∘C0^\circ\text{C} into water at 10∘C10^\circ\text{C}? (Specific latent heat of ice = 336 J g−1336 \text{ J g}^{-1}, Specific heat capacity of water = 4.2 J g−1 K−14.2 \text{ J g}^{-1} \text{ K}^{-1})

Solution:

Step 1: Heat to melt ice (Q1=mLfQ_1 = mL_f) = 10×336=3360 J10 \times 336 = 3360 \text{ J}. Step 2: Heat to raise water temperature (Q2=mcΔTQ_2 = mc\Delta T) = 10×4.2×(10−0)=420 J10 \times 4.2 \times (10 - 0) = 420 \text{ J}. Total heat Q=Q1+Q2=3360+420=3780 JQ = Q_1 + Q_2 = 3360 + 420 = 3780 \text{ J}.

Explanation:

This involves two stages: a phase change (melting) where temperature is constant, followed by a temperature increase in the liquid state.