Question

1 kg of water is heated from 40 C to 70 C, If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J${kg}^{-1}{K}^{-1})$

• A. $2.44\times {10}^5$J
• B. $1.62\times {10}^5$J
• C. $1.24\times {10}^5$J
• D. $2.62\times {10}^5$J

Answer 1

The correct option is C $1.24\times {10}^5$J

Since volume does not change as the energy supplied is used to change the internal energy.

$\Delta E=\Delta U=mC\theta =1\times 4148\times 30=1.24\times {10}^{5}J$