Chapter 16 - Exercises and Problems - Page 310: 35

To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant. In this case, the closed system consists of the frying pan and the water in the sink. The energy transferred from the frying pan to the water is given by the formula: Q = mcpΔT Where Q is the amount of heat transferred, m is the mass of the object, cp is the specific heat capacity, and ΔT is the change in temperature. Let's first calculate the amount of heat transferred from the frying pan to the water: Q1 = m1 * cp1 * ΔT1 Q1 = 2.65 kg * 0.45 kJ/kg°C * (144°C - T) Where cp1 is the specific heat capacity of the frying pan and T is the final temperature of the frying pan and water. The energy gained by the water is: Q2 = m2 * cp2 * ΔT2 Q2 = 10.9 kg * 4.18 kJ/kg°C * (T - 21°C) Where cp2 is the specific heat capacity of water. Since there is no heat loss, the energy gained by the water is equal to the energy lost by the frying pan: Q1 = Q2 2.65 kg * 0.45 kJ/kg°C * (144°C - T) = 10.9 kg * 4.18 kJ/kg°C * (T - 21°C) Simplifying the equation, we get: 0.45 * 2.65 * (144 - T) = 4.18 * 10.9 * (T - 21) Solving for T, we get: T = 34.8°C Therefore, the equilibrium temperature of the water and frying pan is 34.8°C.

To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant. In this case, the closed system consists of the frying pan and the water in the sink. The energy transferred from the frying pan to the water is given by the formula: Q = mcpΔT Where Q is the amount of heat transferred, m is the mass of the object, cp is the specific heat capacity, and ΔT is the change in temperature. Let's first calculate the amount of heat transferred from the frying pan to the water: Q1 = m1 * cp1 * ΔT1 Q1 = 2.65 kg * 0.45 kJ/kg°C * (144°C - T) Where cp1 is the specific heat capacity of the frying pan and T is the final temperature of the frying pan and water. The energy gained by the water is: Q2 = m2 * cp2 * ΔT2 Q2 = 10.9 kg * 4.18 kJ/kg°C * (T - 21°C) Where cp2 is the specific heat capacity of water. Since there is no heat loss, the energy gained by the water is equal to the energy lost by the frying pan: Q1 = Q2 2.65 kg * 0.45 kJ/kg°C * (144°C - T) = 10.9 kg * 4.18 kJ/kg°C * (T - 21°C) Simplifying the equation, we get: 0.45 * 2.65 * (144 - T) = 4.18 * 10.9 * (T - 21) Solving for T, we get: T = 34.8°C Therefore, the equilibrium temperature of the water and frying pan is 34.8°C.