Let $m_1$ be the mass of water at $T_1$, and $m_2$ be the mass of water at $T_2$.
We are given that $m_1 = 40$ g, $T_1 = 40^\circ C$, $m_2 = 10$ g, and $T_2 = 80^\circ C$. Let $T_f$ be the final temperature of the mixture.
We can use the principle of calorimetry, which states that the heat lost by the hotter water is equal to the heat gained by the colder water.
Let $c$ be the specific heat of water.
Heat gained by the colder water is $Q_1 = m_1 c (T_f - T_1) = 40c(T_f - 40)$
Heat lost by the hotter water is $Q_2 = m_2 c (T_2 - T_f) = 10c(80 - T_f)$
According to the principle of calorimetry, $Q_1 = Q_2$, so $40c(T_f - 40) = 10c(80 - T_f)$
Since $c$ is a common factor, we can divide both sides by $10c$: $4(T_f - 40) = 80 - T_f$ $4T_f - 160 = 80 - T_f$ $5T_f = 240$ $T_f = \frac{240}{5} = 48$
So the final temperature of the mixture is $48^\circ C$.
Under the same load, wire A having length $50 m$ and cross section $25 \times 10^{-5} m ^2$ stretches uniformly by the same amount as another wire $B$ of length $60 m$ and a cross section of $30 \times 10^{-5}$ $m ^2$ stretches The ratio of the Young's modulus of wire $A$ to that of wire $B$ will be :
A block of ice of mass $120\, g$ at temperature $0^{\circ} C$ is put in $300\, gm$ of water at $25^{\circ} C$ The $xg$ of ice melts as the temperature of the water reaches $0^{\circ} C$ The value of $x$ is [Use: Specific heat capacity of water $=4200\,Jkg ^{-1} K ^{-1}$, Latent heat of ice = 3.5 \(\times\) 105 JKg-1]