Question

An electric heater of power $600$ W raises the temperature of $4.0$ kg of a liquid from ${10.0}^{0}C$ to ${15.0}^{0}C$ in $100$s. Calculate: (i) The heat capacity of $4.0$ kg of liquid, and (ii) the specific heat capacity of liquid.

• A. (i) $1.2\times {10}^{4}J{K}^{-1}$
  (ii) $30\times {10}^{3}JK{g}^{-1}{K}^{-1}$
• B. (i) $1.2\times {10}^{6}J{K}^{-1}$
  (ii) $3\times {10}^{3}JK{g}^{-1}{K}^{-1}$
• C. (i) $1.2\times {10}^{4}J{K}^{-1}$
  (ii) $3\times {10}^{3}JK{g}^{-1}{K}^{-1}$
• D. (i) $1.2\times {10}^{4}J{K}^{-1}$
  (ii) $3\times {10}^{8}JK{g}^{-1}{K}^{-1}$

Answer 1

Answer: (C)

(i) $1.2\times {10}^{4}J{K}^{-1}$

(ii) $3\times {10}^{3}JK{g}^{-1}{K}^{-1}$
Given:

Power, P = 600W

Mass, m= 4 Kg

Initial temperature, $T_1=10°C=(10+273)K=283K$

Final temperature, $T_2=15°C=(15+273)K=288K$

Time, t=100 s

Let c be the specific heat capacity

Solution:

Heat energy supplied ,Q = mc$\delta T$

$\Rightarrow P\times t=4\times c\times (288-283)$

$\Rightarrow 600\times 100=4\times c\times 5$

$\Rightarrow c=\frac{600\times 100}{4\times 5}$

$\Rightarrow c=3000J/KgK$

Therefore, The specific heat capacity, c = $3\times {10}^{3}J/KgK$

Heat capacity , S = mc

$\Rightarrow S=4\times 3000J/K$

$\Rightarrow S=12000J/K$

Therefore, Heat capacity ,S = 1.2 ×${10}^{4}J/K$

Hence, Option C is correct option.