Water of mass 120g at 50oC is added to 200g of water at 10oC and the mixture is well stirred. Calculate the temperature of the mixture. [Neglect heat losses...

To solve this problem, we need to use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. When the hot water is mixed with the cold water, heat energy flows from the hot water to the cold water until they reach the same temperature. We can use the equation: Q = mCΔT where Q is the amount of heat energy transferred, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the change in temperature. Let's start by finding the amount of heat energy transferred from the hot water to the cold water: Q₁ = (0.12 kg) * (4186 J/kg°C) * (50°C - T) where T is the final temperature of the mixture. We use 4186 J/kg°C as the specific heat capacity of water. Similarly, the amount of heat energy gained by the cold water is: Q₂ = (0.2 kg) * (4186 J/kg°C) * (T - 10°C) Since energy is conserved, we know that Q₁ = Q₂. Therefore: (0.12 kg) * (4186 J/kg°C) * (50°C - T) = (0.2 kg) * (4186 J/kg°C) * (T - 10°C) Solving for T, we get: T = 25°C Therefore, the temperature of the mixture is 25°C. Option (C) is correct. Note: It is important to keep in mind that this calculation neglects any heat losses to the surroundings, which may affect the actual temperature of the mixture in a real-world scenario.