Question

800 calories of heat is required to raise the temperature of 0.080kg of a liquid from 10°C to 100°C. Find its specific heat capacity in cal /kg°C

Answer 1

Answer:

111.1 cal / kg C

Explanation:

Given :

Given heat Q = 800 cal

Mass m = 0.080 kg

We have temperature difference Δ T = 100 - 10 C

Δ T = 90 C

We know :

Q = m c Δ T

c = Q / m Δ T    [ where c = specific heat ]

c = ( 800 ) / ( 0.080 × 90 ) cal / kg C

c = 800 / 7.2 cal / kg C

c =  111.1 cal / kg C

Therefore , specific heat capacity of liquid is  111.1 cal / kg C

Answer 2

Answer:

• Specific heat capacity (c) of liquid is 111.1 cal / Kg °C.

Given:

1. Heat supplied (Q) = 800 calories.
2. Mass of the liquid (M) = 0.080 Kg.
3. Initial temperature (T_i) = 10°C
4. Final Temperature (T_f) = 100°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,

⇒ 800 = 0.080 × c × ΔT

⇒ 800 = 0.080 × c × (T_f - T_i)      ∵[ΔT = T_f - T_i]

⇒ 800 = 0.080 × c × (100 - 10)

⇒ 800 = 0.080 × c × 90

⇒ 800 = 7.2 × c

⇒ c = 800 / 7.2

⇒ c = 111.11

⇒ c = 111.11 cal / Kg °C

∴ Specific heat capacity (c) of liquid is 111.1 cal / Kg °C

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