Question

1 An electric heater of power 600 W raises the
temperature of 4.0 kg of a liquid from 10:0°C to 15.0 °C in 100 s. Calculate :
(i) the heat capacity of 4.0 kg of liquid, and
(ii) the specific heat capacity of liquid.

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Answer 1

GivEn :

• Power of heater (P) = 600 W

• Mass of the liquid (m) = 4.0 kg

• Time taken = 100s

• Change in temperature = 15° - 10° = 5° C

To find :

• Heat capacity of 4.0 kg of liquid = ?

• Specific heat capacity of liquid = ?

SoluTion :

To solve the above question, understand the concept of Specific Heat !!

Specific heat is the total amount of heat per unit mass required to raise the temperature by 1° C.

It is given by,

${\large{ \dag \: { \boxed{ \tt{ \blue{Q = mc \triangle \: T}}}}}}$

where,

• Q = heat energy

• m = mass of the object

• c = specific heat

Also, power in terms of specific heat can be given as ,

${\large{ \dag \: { \boxed{ \tt{ \blue{\triangle Q = p \times t }}}}}}$

where,

• p = power

• t = required time

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According to the above formulas,

$\tt \dashrightarrow \: \: \: \triangle Q = P \times t$

$\tt \dashrightarrow \: \: \: \triangle Q =600 \times 100$

$\tt \dashrightarrow \: \: \: {\underline{\boxed{\orange{\tt{ \triangle Q = 60000 J}}}}}$

Now,

$\tt \dashrightarrow \: \: \: \triangle Q = mc \triangle \: T$

$\tt \dashrightarrow \: \: \: c = \dfrac{ \triangle Q }{m \times \triangle T}$

$\tt \dashrightarrow \: \: \:c = \dfrac{60000}{4 \times 5}$

$\tt \dashrightarrow \: \: \: {\underline{\boxed{\red{\tt{c = 3000 J kg/k}}}}}$

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Specific Heat capacity of the liquid,

$\tt \dashrightarrow \: \: \: Heat \; capacity = c \times m$

$\tt \dashrightarrow \: \: \: Heat \; capacity = 4 \times 3000 J kg/k$

$\tt \dashrightarrow \: \: \:Heat \; capacity = 12000 J kg/k$

$\tt \dashrightarrow \: \: \: {\underline{\boxed{\purple{\tt{ Heat \; capacity = 1.2 \times 10 J/K}}}}} \; \; \; \; [After \; conversion]$

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