Question

A $100$ gm of water is stirred and the work done during this process is $4200$ J. The rise in temperature during the process would be :

• A. ${0.01}^{o}C$
• B. ${0.1}^{o}C$
• C. $1^{o}C$
• D. ${10}^{o}C$

Answer 1

The correct option is D ${10}^{o}C$

The energy absorbed by water in stirring is given by,
$H=mC\Delta T$
Substituting the values, m = 0.1 kg, C = 4200 J/kg-K and H = 4200 J
This gives us, $\Delta T={10}^{o}C$