Question

10 g of water at 70^@C is mixed with 5 g of water at 30^@C. Find the temperature of the mixture in equilibrium. Specific heat of water is 1 cal//g-^@C.

Answer 1

The temperature of the mixture in equilibrium is

Explanation:

Given :

Mass of water $m _{1} = 10$ g

Temperature of first water $T_{1} =$ 70°C

Mass of another water $m_{2} = 5$ g

Temperature of second water $T_{2} =$ 30°C

Specific heat of water $= 1 \frac{cal}{g C}$

From the conservation of energy,

⇒ Heat given by first water = Heat taken by second water

   $m_{1} C \Delta T = m_{2} C \Delta T$

   $10 \times 1 \times (70 - T_{o} ) = 5 \times \times 1 \times (T_{o} -30 )$

Solving above equation we get,

   $T _{o } =$ 56.7°C

Thus, the temperature of mixture at equilibrium is 56.7°C