A vessel contains \(110 \mathrm{~g}\) of water the heat capacity of the vessel is equal to \(10 \mathrm{~g}\) of water. The initial temperature of water in vessel is \(10^{\circ} \mathrm{C}\) If \(220 \mathrm{~g}\) of hot water at \(70^{\circ} \mathrm{C}\) is poured in the vessel the Final temperature neglecting radiation loss will be (A) \(70^{\circ} \mathrm{C}\) (B) \(80^{\circ} \mathrm{C}\) (C) \(60^{\circ} \mathrm{C}\) (D) \(50^{\circ} \mathrm{C}\)

As we stated earlier, heat gained by the cooler water is equal to the heat lost by the hotter water. We can set up an equation for this. \(Q_{gained} = Q_{lost}\)

Now we will define the variables for the masses and temperatures of water: \(m_1 = 110\) g (mass of the cooler water) \(m_2 = 220\) g (mass of the hotter water) \(T_1 = 10\) °C (initial temperature of the cooler water) \(T_2 = 70\) °C (initial temperature of the hotter water) \(C = 1\) cal/g·°C (specific heat capacity of water) \(V = 10\) g (the heat capacity of the vessel is equal to this) Now, the equation for heat transfer will involve the specific heat capacity of water.

Heat gained by the cooler water (including the vessel) can be expressed as: \(Q_{gained} = (m_1 + V) * C * (T_f - T_1)\)

Heat lost by the hotter water can be expressed as: \(Q_{lost} = m_2 * C * (T_2 - T_f)\)

Now we can set up the equation with all gathered data. \((m_1 + V) * C * (T_f - T_1) = m_2 * C * (T_2 - T_f)\) As \(C\) is present on both sides of the equation, it can be cancelled out. The equation becomes: \((110 + 10) (T_f - 10) = 220 (70 - T_f)\) Solving for \(T_f\): \(120(T_f - 10) = 220(70 - T_f)\) Now, expand and simplify the equation: \(120T_f - 1200 = 15400 - 220T_f\) Combine like terms: \(340T_f = 16600\) Now, divide both sides by 340: \(T_f = 48.8235294\) Since we need the final temperature in whole numbers, we can round up the final value to the nearest whole number \(T_f \approx 49^{\circ} \mathrm{C}\) The final temperature is between \(50^{\circ} \mathrm{C}\) and \(60^{\circ} \mathrm{C}\), which means our answer is closest to option (D) \(50^{\circ} \mathrm{C}\).