Question

the heat capacity of $4.0kg$ of liquid :

• A. $0.9\times {10}^{4}J{K}^{-1}$
• B. $0.6\times {10}^{4}J{K}^{-1}$
• C. $1.2\times {10}^{4}J{K}^{-1}$
• D. $2.4\times {10}^{4}J{K}^{-1}$

Answer 1

Answer: (C)

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

Heat supplied by heater is $600\times 100=60000J$

So heat capacity is $\frac{E}{\Delta T}$

$\frac{60000}{(15-10)}=\frac{60000}{5}=1.2\times {10}^{4}J{K}^{-1}$