Question

Initially a beaker had $100$ g of water at temperature $90^\circ$C. Later another $600$ g of water at temperature $20^\circ$C was poured into the beaker. The temperature, $T$, of the water after mixing is

• A. $20^\circ$C
• B. $30^\circ$C
• C. $45^\circ$C
• D. $55^\circ$C
• E. $90^\circ$C

Hint:

For same substance, specific heat cancels out in heat balance equations.

Answer 1

The Correct Option is B

Concept:
Heat lost = Heat gained: \[ m_1 c (T_1 - T) = m_2 c (T - T_2) \]

Step 1: Substitute values.
\[ 100(90 - T) = 600(T - 20) \]

Step 2: Solve.
\[ 9000 - 100T = 600T - 12000 \] \[ 21000 = 700T \Rightarrow T = 30^\circ C \]