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Heat - Heat and Temperature

Grade 7ICSE

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 SI unit is the Joule (JJ), and another common unit is the calorie (calcal), where 1 cal≈4.186 J1 \text{ cal} \approx 4.186 \text{ J}.

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Temperature is the measure of the degree of hotness or coldness of a body. The SI unit of temperature is Kelvin (KK), though Celsius (∘C^\circ C) and Fahrenheit (∘F^\circ F) are commonly used in daily life.

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The Celsius scale has a lower fixed point (melting point of ice) at 0∘C0^\circ C and an upper fixed point (boiling point of water) at 100∘C100^\circ C.

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The Fahrenheit scale has a lower fixed point at 32∘F32^\circ F and an upper fixed point at 212∘F212^\circ F.

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The Kelvin scale is the absolute scale of temperature. The relationship between Kelvin and Celsius is K=C+273K = C + 273. Absolute zero is 0 K0 \text{ K} or −273∘C-273^\circ C.

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Thermal expansion is the increase in the size of an object when heated. Different materials expand by different amounts for the same rise in temperature.

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A Clinical Thermometer is used to measure human body temperature, ranging from 35∘C35^\circ C to 42∘C42^\circ C. It has a 'kink' to prevent the backflow of mercury.

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Specific Heat Capacity is the amount of heat energy required to raise the temperature of a unit mass of a substance by 1∘C1^\circ C. It is denoted as ss or cc.

📐Formulae

C5=F−329\frac{C}{5} = \frac{F - 32}{9}

K=C+273K = C + 273

Q=m×s×ΔTQ = m \times s \times \Delta T

1 kcal=1000 calories=4186 J1 \text{ kcal} = 1000 \text{ calories} = 4186 \text{ J}

💡Examples

Problem 1:

The normal temperature of the human body is 98.6∘F98.6^\circ F. Convert this temperature into the Celsius scale (∘C^\circ C).

Solution:

Using the formula C5=F−329\frac{C}{5} = \frac{F - 32}{9}, substitute F=98.6F = 98.6: C5=98.6−329\frac{C}{5} = \frac{98.6 - 32}{9} C5=66.69\frac{C}{5} = \frac{66.6}{9} C=7.4×5=37∘CC = 7.4 \times 5 = 37^\circ C

Explanation:

By applying the conversion formula, we find that the standard human body temperature in Celsius is 37∘C37^\circ C.

Problem 2:

Convert a room temperature of 25∘C25^\circ C to the Kelvin scale.

Solution:

Using the formula K=C+273K = C + 273: K=25+273K = 25 + 273 K=298 KK = 298 \text{ K}

Explanation:

To convert Celsius to Kelvin, we add 273273 to the Celsius value because the Kelvin scale starts at absolute zero (−273∘C-273^\circ C).

Problem 3:

Calculate the amount of heat required to raise the temperature of 2 kg2 \text{ kg} of water from 20∘C20^\circ C to 60∘C60^\circ C. (Take specific heat of water s=4200 J/kg∘Cs = 4200 \text{ J/kg}^\circ C)

Solution:

Given m=2 kgm = 2 \text{ kg}, s=4200 J/kg∘Cs = 4200 \text{ J/kg}^\circ C, and ΔT=60∘C−20∘C=40∘C\Delta T = 60^\circ C - 20^\circ C = 40^\circ C. Using Q=m×s×ΔTQ = m \times s \times \Delta T: Q=2×4200×40Q = 2 \times 4200 \times 40 Q=336,000 J or 336 kJQ = 336,000 \text{ J} \text{ or } 336 \text{ kJ}

Explanation:

Heat energy (QQ) is the product of mass, specific heat capacity, and the change in temperature (ΔT\Delta T).