Question

Initially, a beaker has $100\,g$ 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

Answer 1

Given,
Mass of water at $90^\circ C=100\,g$
$=100\times 10^{- 3}kg$
Mass of water at $20^\circ C=600 \, g$
$=600\times 10^{- 3}kg$
From calorimetry,
$\left(100 \times \left(10\right)^{- 3}\right)S_{w}\left(\right. 90 ^\circ C - T ^\circ C \left.\right)=\left(600 \times \left(10\right)^{- 3}\right)S_{w}\left(\right. T ^\circ C - 20 ^\circ C \left.\right)$
$\Rightarrow 90-T=6T-120$
$T=\frac{210}{7}=30^\circ C$