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Energy - Specific Heat

Grade 9IB

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

🔑Concepts

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Specific Heat Capacity (cc) is defined as the amount of heat energy required to raise the temperature of 1 kg1\ kg of a substance by 1∘C1^{\circ}C (or 1 K1\ K).

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Heat energy (QQ) is a form of energy transfer between systems due to a temperature difference, measured in Joules (JJ).

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Temperature (TT) is a measure of the average kinetic energy of the particles in a substance.

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The change in thermal energy depends on the mass (mm) of the object, the specific heat capacity (cc) of the material, and the change in temperature (ΔT\Delta T).

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Substances with a high specific heat capacity, such as water (c≈4200 J/kg∘Cc \approx 4200\ J/kg^{\circ}C), require more energy to change their temperature compared to substances with low specific heat capacity, like metals.

📐Formulae

Q=mcΔTQ = m c \Delta T

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

c=QmΔTc = \frac{Q}{m \Delta T}

💡Examples

Problem 1:

Calculate the amount of heat energy required to raise the temperature of 0.5 kg0.5\ kg of water from 25∘C25^{\circ}C to 75∘C75^{\circ}C. The specific heat capacity of water is 4200 J/kg∘C4200\ J/kg^{\circ}C.

Solution:

Given: m=0.5 kgm = 0.5\ kg c=4200 J/kg∘Cc = 4200\ J/kg^{\circ}C Tinitial=25∘CT_{initial} = 25^{\circ}C Tfinal=75∘CT_{final} = 75^{\circ}C

Step 1: Calculate the change in temperature: ΔT=75−25=50∘C\Delta T = 75 - 25 = 50^{\circ}C

Step 2: Use the heat energy formula: Q=mcΔTQ = m c \Delta T Q=0.5×4200×50Q = 0.5 \times 4200 \times 50 Q=2100×50Q = 2100 \times 50 Q=105000 JQ = 105000\ J

Explanation:

To find the total energy, we multiply the mass by the specific heat capacity and then by the temperature difference. The result 105000 J105000\ J can also be written as 105 kJ105\ kJ.

Problem 2:

A 2 kg2\ kg block of metal absorbs 18000 J18000\ J of energy, causing its temperature to rise by 20∘C20^{\circ}C. Find the specific heat capacity of the metal.

Solution:

Given: m=2 kgm = 2\ kg Q=18000 JQ = 18000\ J ΔT=20∘C\Delta T = 20^{\circ}C

Step 1: Rearrange the formula to solve for cc: c=QmΔTc = \frac{Q}{m \Delta T}

Step 2: Substitute the values: c=180002×20c = \frac{18000}{2 \times 20} c=1800040c = \frac{18000}{40} c=450 J/kg∘Cc = 450\ J/kg^{\circ}C

Explanation:

By dividing the total energy absorbed by the product of mass and temperature change, we determine that the metal has a specific heat capacity of 450 J/kg∘C450\ J/kg^{\circ}C, which is typical for metals like iron or steel.