Question

Calculate the specific heat capacity of a substance if it takes
542000J of energy to heat up 1kg of it from 40 to 60°C

Answer 1

Answer:

• Specific heat capacity (c) of liquid is 2710 J / Kg °C.

Given:

1. Heat supplied (Q) = 542000 J.
2. Mass of the liquid (M) = 1 Kg.
3. Initial temperature (T_i) = 40°C
4. Final Temperature (T_f) = 60°C.

Explanation:

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Specific Heat capacity (c):

The amount of heat required to raise the temperature of unit of substance ( solid / liquid / gases) by unity i.e. 1 Joule/calorie is known as specific heat capacity of a substance.

It is denoted by " c ".

Units:

• S.I unit = J Kg⁻¹ K⁻¹
• C.G.S unit = Cal g⁻¹ K⁻¹

Dimensional Formula:

• M⁰ L² T⁻² K⁻¹

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From the formula we know,

⇒ Q = M c ΔT

Where,

• Q denotes Heat energy.
• M denotes mass.
• ΔT denotes temperature.

Now,

⇒ Q = M c ΔT

Substituting the values,

⇒ 542000 = 1 × c × ΔT

⇒ 542000 = 1 × c × (T_f - T_i)      ∵[ΔT = T_f - T_i]

⇒ 542000 = 1 × c × (60 - 40)

⇒ 542000 = 1 × c × 20

⇒ 542000 = 20 × c

⇒ c = 542000 / 20

⇒ c = 54200 / 2

⇒ c = 27100

⇒ c = 27100 J / Kg °C

∴ Specific heat capacity (c) of liquid is 27100 J / Kg °C

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