Water of mass 150g at \(60^\circ\)c is added to 300g of water at \(20^\circ\)c and the mixture is well stirred. Calculate the temperature of the mixture.(ne...

Answer Details

To solve this problem, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the energy transferred is in the form of heat. We can use the formula: Q = m*c*(ΔT) where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, we can calculate the heat transferred from the hot water to the cold water: Q1 = 150g * 4.18 J/(g°C) * (60°C - T) Q1 = 627 * (60 - T) where T is the temperature of the mixture. Next, we can calculate the heat transferred from the cold water to reach the final temperature of the mixture: Q2 = 300g * 4.18 J/(g°C) * (T - 20°C) Q2 = 1254 * (T - 20) Since the heat transferred between the two water samples must be equal, we can set Q1 equal to Q2 and solve for T: 627 * (60 - T) = 1254 * (T - 20) 37620 - 627T = 1254T - 25080 1881T = 62760 T = 33.4°C Therefore, the temperature of the mixture is approximately 33°C. Answer: 33°C