A piece of copper of mass 400g at the temperature of 1000°c is quickly transferred to a vessel of negligible thermal capacity containing 300g of water at 29°c if the final steady temperature of the mixture is 108°c calculate the mass of the Water that boils away.
Final temp of water/copper is 108C ? Then all the water must have boiled away. max temp of water is 100C
I need answer
To find the mass of the water that boils away, we need to understand the concept of heat transfer and use the specific heat capacity of different substances.
First, let's calculate the heat gained by the copper and the heat lost by the water during the process:
1. Heat gained by the copper (Qcopper)
Qcopper = mass_copper * specific_heat_copper * temperature_change_copper
In this case, mass_copper = 400g
specific_heat_copper = 0.385 J/g°C (specific heat capacity of copper)
temperature_change_copper = final temperature - initial temperature
= 108°C - 1000°C
2. Heat lost by the water (Qwater)
Qwater = mass_water * specific_heat_water * temperature_change_water
In this case, mass_water = 300g
specific_heat_water = 4.184 J/g°C (specific heat capacity of water)
temperature_change_water = final temperature - initial temperature
= 108°C - 29°C
Now, assuming no heat loss to the surroundings (negligible thermal capacity of the vessel), the heat gained by the copper is equal to the heat lost by the water:
Qcopper = Qwater
Next, rearrange the equation and solve for the mass of water that boils away (let's call it mass_boiled_water):
mass_boiled_water = (Qcopper - Qwater) / latent_heat_vaporization_water
Here, the latent heat of vaporization of water is 2260 J/g.
Substituting the values into the equation, we can calculate the mass of water that boils away.
Please note that the specific heat capacity and latent heat values for copper and water are approximate averages and may vary slightly.