Question

Two taps supply water to a container, one at the temperature of \(20^\circ\)C at the rate of \(2\) kg/minute and another at \(80^\circ\)C at the rate of \(1\) kg/minute. If the container gets water from the taps simultaneously for \(10\) minutes, then the temperature of water in the container is

• A. \(35^\circ\)C
• B. \(30^\circ\)C
• C. \(50^\circ\)C
• D. \(40^\circ\)C
• E. \(45^\circ\)C

Hint:

For mixing water with no heat loss: \[ T=\frac{\sum m_iT_i}{\sum m_i} \]

Answer 1

The Correct Option is D

Masses entering in \(10\) minutes: From first tap: \[ m_1=2\times 10=20\text{ kg} \] From second tap: \[ m_2=1\times 10=10\text{ kg} \] Final temperature: \[ T=\frac{m_1T_1+m_2T_2}{m_1+m_2} \] \[ T=\frac{20(20)+10(80)}{20+10} \] \[ T=\frac{400+800}{30} =\frac{1200}{30}=40^\circ\text{C} \] Hence, \[ \boxed{(D)\ 40^\circ\text{C}} \]