a piece of copper of mass 300g at 950 degrees celcius is quickly transfered to vessel of negligible thermole capacity containing 250g of water at 25 degrees celcious. if the final steady temp of the mixture is 100degrees, calculate the masd of water that will boil away
The copper is cooled from 950 to 250
so heat out of copper = 300 *Ccopper (950 -100)
that heat goes into the water which of course ends up at 100 deg if boiling.
heat into water = 250 Cwater (100 -25) + Mass boiled* heat of vaporization of water
solve for Mass boiled.
what is this,you are very stupid,where is your answer
this is rubbish
To calculate the mass of water that will boil away, we can use the concept of heat transfer and assume no heat is lost to the surroundings.
First, let's calculate the initial heat energy of the copper and the final heat energy of the mixture. We'll use the specific heat capacity (c) of copper and water in the calculations.
1. Calculate the initial heat energy of the copper:
Q1 = mass of copper × specific heat capacity of copper × change in temperature
= 300 g × specific heat capacity of copper × (final temperature - initial temperature)
Since the initial temperature of the copper is 950 degrees Celsius, and the specific heat capacity of copper is about 0.39 J/g°C, the formula becomes:
Q1 = 300 g × 0.39 J/g°C × (100°C - 950°C)
2. Calculate the final heat energy of the mixture:
Q2 = (mass of copper + mass of water) × specific heat capacity of water × (final temperature - initial temperature)
Given the mass of water is 250 g, the specific heat capacity of water is about 4.18 J/g°C, and the initial temperature of water is 25 degrees Celsius, the formula becomes:
Q2 = (300 g + 250 g) × 4.18 J/g°C × (100°C - 25°C)
Now, as no heat is lost to the surroundings, the initial heat energy of the copper must be equal to the final heat energy of the mixture:
Q1 = Q2
Plugging in the respective values, we get:
300 g × 0.39 J/g°C × (100°C - 950°C) = (300 g + 250 g) × 4.18 J/g°C × (100°C - 25°C)
Now, let's solve this equation to find the mass of water that will boil away.
(Note: It's important to note that as the copper cools down to reach the final temperature of 100 degrees Celsius, it transfers heat energy to the water, causing some of the water to boil away. The mass of water that boils away will be the difference between the initial mass of water and the final mass of water.)
Let's denote the mass of water that boils away as 'm':
Initial mass of water - Final mass of water = m
Using the equation we derived earlier, we can rearrange it to solve for 'm':
m = Initial mass of water - (Q1 / (specific heat capacity of water × ΔT))
Now, plug in the values to calculate the mass of water that boils away.