Question

A 49.23 g red hot piece of steel (specific heat capacity = 0.466 J/g· °C) at a temperature of 649 °C is dropped into 984 g of water that is initially at a temperature of 11.2 °C. What is the final temperature of the system, in °C, once it reaches equilibrium? Assume no energy loss to the environment.

Answer 1

To determine the final equilibrium temperature when a hot piece of steel is dropped into water, we set the heat lost by steel equal to the heat gained by water using the equation Q = m × c × ∆T and solve for the final temperature.

Step-by-step explanation:

To find the final temperature of the system after a 49.23 g piece of steel at 649 °C is dropped into 984 g of water at 11.2 °C, we can use the principle of conservation of energy. We assume no energy is lost to the environment, meaning the heat lost by the steel will equal the heat gained by the water.

The heat change for the steel can be described by the equation
Q = m × c × ∆T, where m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. For water, the specific heat capacity ( c_water ) is 4.18 J/g°C.

Let T_final be the final temperature for both steel and water. For steel, the heat change is
(49.23 g) × (0.466 J/g°C) × (649 °C - T_final). For water, the heat change is (984 g) × (4.18 J/g°C) × (T_final - 11.2 °C).

Setting the heat lost by steel equal to the heat gained by water and solving for T_final gives us the final equilibrium temperature:

(49.23 g) × (0.466 J/g°C) × (649 °C - T_final) = (984 g) × (4.18 J/g°C) × (T_final - 11.2 °C)