What is the final equilibrium temperature when a 0.400 kg iron horseshoe at 500°C is dropped into 20 kg of water at 22°C?

To determine the final equilibrium temperature when the iron horseshoe is dropped into the water, we can apply the principle of conservation of energy. Here, the heat lost by the iron will equal the heat gained by the water, assuming no heat is lost to the surroundings.

The specific heat capacity of iron is around 450 J/(kg·°C), while that of water is approximately 4186 J/(kg·°C). Using these values, we can set up the energy balance equation:

Heat lost by iron = Heat gained by water

We denote the final equilibrium temperature as ( T_f ).

For the iron horseshoe:

• Mass = 0.400 kg

• Initial temperature = 500°C

• Final temperature = ( T_f )

The heat loss can be expressed as:

[ Q_{iron} = m_{iron} \cdot c_{iron} \cdot (T_{initial} - T_f) ]

[ Q_{iron} = 0.400 , \text{kg} \cdot 450 , \text{J/(kg·°C)} \cdot (500°C - T_f) ]

For the water:

• Mass = 20 kg