A piece of copper of mass 300 g at a temperature of 950 degree Celsius is quickly transferred to a vessel of negligible thermal containing 250 g of water at 25 degree Celsius. If the final temperature of the mixture is 100 degree Celsius calculate the mass of the water that will boil away ( specific heat capacity of copper =400 j/kg.k specific latent heat of vaporization of steam =2260000jkg

0.300 * 400 * (950-100) = 102,000 Joules lost by copper

that heats 0.250 kg water to 100 from 25
0.250 * 4186 * 75 = 78488 Joules used from copper to heat water.
That leaves 23,512 Joules to boil water
23,512 = m 2,260,000
m = .0104 kg = 10.4 grams

What if the special heat capacity of water is 4.2×10^2

Where is the 4186 from

Thanks

Where is the 4186 from? i am confused

Yes

That not the answer

i dont even understand this solving sef

I don't understand the solving

To calculate the mass of water that will boil away, we need to determine the amount of heat transferred from the copper to the water. Considering the following steps:

1. Calculate the heat transferred from the copper to the water using the formula:
Q1 = mcΔT1
Where:
- Q1 is the heat transferred from the copper (in Joules)
- m is the mass of the copper (in kg)
- c is the specific heat capacity of copper (in J/kg·K)
- ΔT1 is the change in temperature of the copper (final temperature - initial temperature)

Given values:
- m = 0.3 kg (mass of the copper)
- c = 400 J/kg·K (specific heat capacity of copper)
- ΔT1 = 100°C - 950°C = -850°C (change in temperature of the copper)

Plugging in the values:
Q1 = 0.3 kg × 400 J/kg·K × -850°C

2. Calculate the heat required to raise the temperature of the water from 25°C to 100°C:
Q2 = mwCwΔT2
Where:
- Q2 is the heat required to raise the temperature of the water (in Joules)
- mw is the mass of the water (in kg)
- Cw is the specific heat capacity of water (in J/kg·K)
- ΔT2 is the change in temperature of the water (final temperature - initial temperature)

Given values:
- mw = 0.25 kg (mass of the water)
- Cw = 4186 J/kg·K (specific heat capacity of water)
- ΔT2 = 100°C - 25°C = 75°C (change in temperature of the water)

Plugging in the values:
Q2 = 0.25 kg × 4186 J/kg·K × 75°C

3. Calculate the heat required to boil away the mass of water:
Q3 = mlv
Where:
- Q3 is the heat required to boil away the water (in Joules)
- ml is the mass of water to be boiled away (unknown)
- lv is the specific latent heat of vaporization of water (in J/kg)

Given values:
- lv = 2260000 J/kg (specific latent heat of vaporization of steam)

4. The total heat transferred from the copper to the water is equal to the heat required to raise the temperature of the water plus the heat required to boil away the water:
Q1 + Q2 = Q3

Plugging in the calculated values and variables:
(0.3 kg × 400 J/kg·K × -850°C) + (0.25 kg × 4186 J/kg·K × 75°C) = ml × 2260000 J/kg

Solve the equation for ml to find the mass of water that will boil away.