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Heat - Effects of Heat

Grade 7ICSE

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

🔑Concepts

Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature. Its SI unit is the Joule (JJ), but it is often measured in calories (calcal).

Change in Temperature: When a body absorbs heat, its average kinetic energy increases, leading to a rise in temperature (TT). Conversely, losing heat leads to a fall in temperature.

Change in State: Heat can cause a substance to change from solid to liquid (melting), liquid to gas (vaporization), or solid directly to gas (sublimation). During these changes, the temperature remains constant.

Thermal Expansion: Most substances expand when heated and contract when cooled. Expansion is greatest in gases, followed by liquids, and least in solids.

Anomalous Expansion of Water: Water behaves uniquely; it contracts when heated from 0C0^{\circ}C to 4C4^{\circ}C and expands when cooled from 4C4^{\circ}C to 0C0^{\circ}C. The density of water is maximum at 4C4^{\circ}C.

Chemical Change: Heat energy can initiate chemical reactions. For example, heating Potassium Chlorate (KClO3KClO_3) in the presence of Manganese Dioxide (MnO2MnO_2) causes it to decompose into Potassium Chloride (KClKCl) and Oxygen (O2O_2).

📐Formulae

1 calorie4.186 J1\text{ calorie} \approx 4.186\text{ J}

C5=F329\frac{C}{5} = \frac{F - 32}{9}

K=C+273K = C + 273

Q=m×s×ΔtQ = m \times s \times \Delta t

💡Examples

Problem 1:

A clinical thermometer reads the body temperature as 104F104^{\circ}F. Calculate this temperature in the Celsius (C^{\circ}C) scale.

Solution:

Using the formula: C5=F329\frac{C}{5} = \frac{F - 32}{9}. Substitute F=104F = 104: C5=104329C5=729C5=8\frac{C}{5} = \frac{104 - 32}{9} \Rightarrow \frac{C}{5} = \frac{72}{9} \Rightarrow \frac{C}{5} = 8. Therefore, C=8×5=40CC = 8 \times 5 = 40^{\circ}C.

Explanation:

To convert Fahrenheit to Celsius, subtract 3232 from the Fahrenheit value, multiply by 55, and then divide by 99.

Problem 2:

Calculate the amount of heat energy required to raise the temperature of 200 g200\text{ g} of water from 20C20^{\circ}C to 30C30^{\circ}C. (Specific heat of water s=1 cal/gCs = 1\text{ cal/g}^{\circ}C)

Solution:

Given m=200 gm = 200\text{ g}, s=1 cal/gCs = 1\text{ cal/g}^{\circ}C, and Δt=(3020)=10C\Delta t = (30 - 20) = 10^{\circ}C. Heat Q=m×s×Δt=200×1×10=2000 caloriesQ = m \times s \times \Delta t = 200 \times 1 \times 10 = 2000\text{ calories}. In Joules, Q=2000×4.2=8400 JQ = 2000 \times 4.2 = 8400\text{ J}.

Explanation:

The heat required is the product of the mass of the substance, its specific heat capacity, and the change in temperature.

Effects of Heat - Revision Notes & Key Formulas | ICSE Class 7 Science