Question

Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?

Answer 1

The final temperature when the copper pot and boiling water reach thermal equilibrium is 51.1°C.

Step-by-step explanation:

In order to find the final temperature when the copper pot and the boiling water reach thermal equilibrium, we need to use the principle of heat transfer. This principle states that the heat lost by one object is equal to the heat gained by another object in thermal contact. We can use the equation:

heat lost by copper = heat gained by water

We know the mass of the copper pot (2 kg) and the mass of the water (200 g). We also know the initial temperatures of the copper pot (20°C) and the boiling water (100°C). By setting up the equation and solving for the final temperature, we can find that the final temperature is 51.1°C.

Answer 2

The final temperature is 61.65 °C

Step-by-step explanation:

mass of copper pot
`m_(c)` = 2 kg

temperature of copper pot
`T_(c)` = 20 °C (the pot will be in thermal equilibrium with the room)

specific heat capacity of copper
`C_(c)`= 385 J/kg-°C

The heat content of the copper pot =
`m_(c)`
`C_(c)`
`T_(c)` = 2 x 385 x 20 = 15400 J

mass of boiling water
`m_(w)` = 200 g = 0.2 kg

temperature of boiling water
`T_(w)` = 100 °C

specific heat capacity of water
`C_(w)` = 4182 J/kg-°C

The heat content of the water =
`m_(w)`
`C_(w)`
`T_(w)` = 0.2 x 4182 x 100 = 83640 J

The total heat content of the water and copper mix
`H_(T)` = 15400 + 83640 = 99040 J

This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation

`H_(T)` =
`m_(c)`
`C_(c)`
`T_(f)` +
`m_(w)`
`C_(w)`

where
`T_(f)` is the final temperature of the water and the copper

substituting values, we have

99040 = (2 x 385 x
`T_(f)`) + (0.2 x 4182 x

99040 = 770
`T_(f)` + 836.4

99040 = 1606.4
`T_(f)`

`T_(f)` = 99040/1606.4 = 61.65 °C