Show that 100 g of 100°C steam will completely melt
800 g of 0°C ice.
Well, that's quite the steamy situation you've got there! But fear not, because I, the Clown Bot, am here to assist you with a side of humor.
To tackle this icy-hot problem, we need to first understand some basic concepts of energy exchange. When the steam comes into contact with the ice, it releases its latent heat, which is the amount of energy required to convert a substance from one state to another without changing its temperature.
Now, let's have some fun with numbers. We know that 100 g of steam at 100°C possesses a certain amount of latent heat. On the other hand, the 800 g of ice at 0°C must absorb this latent heat to melt completely. Since I love exaggerating things, let's imagine that the latent heat exchange is like a dramatic love story where the steam falls head over heels for the ice.
In the process of melting, the steam whispers romantic words like, "You make me melt, baby!" to the ice. The 100 g of steam, being a total charmer, showers the 800 g of ice with enough love and energy to melt it completely. It's like the steam saying, "I'll make you melt for me, ice, whether you like it or not!"
So, to sum it all up, the 100 g of 100°C steam will make the 800 g of 0°C ice melt completely through the exchange of latent heat. It's a steamy love story with a meltingly good ending!
Remember, though, this is just an imaginary scenario. In reality, it's the laws of thermodynamics that govern heat transfer, not love stories. So, don't take my humorous approach too literally!
To show that 100 g of 100°C steam will completely melt 800 g of 0°C ice, we need to calculate the heat gained by the ice to melt it and the heat lost by the steam to cool it down.
Step 1: Calculate the heat gained by the ice to reach its melting point:
The specific heat capacity of ice (cice) is 2.09 J/g°C.
The change in temperature for the ice (ΔTice) is 0°C - (-10°C) = 10°C (since ice starts at -10°C to be raised to its melting point)
The heat gained by the ice (Qice) can be calculated using the formula:
Qice = mass (mice) * specific heat capacity (cice) * ΔTice
Qice = 800 g * 2.09 J/g°C * 10°C
Step 2: Calculate the heat lost by the steam to reach its condensation point:
The specific heat capacity of steam (csteam) is 2.03 J/g°C.
The change in temperature for the steam (ΔTsteam) is 100°C - 100°C = 0°C (since the steam remains at its boiling point)
The heat lost by the steam (Qsteam) can be calculated using the formula:
Qsteam = mass (msteam) * specific heat capacity (csteam) * ΔTsteam
Qsteam = 100 g * 2.03 J/g°C * 0°C
Step 3: Compare Qice and Qsteam to determine if the ice will completely melt:
If Qice is greater than or equal to Qsteam, then the ice will completely melt.
Comparing the two values calculated:
Qice = 800 g * 2.09 J/g°C * 10°C = 16,720 J
Qsteam = 100 g * 2.03 J/g°C * 0°C = 0 J
Since Qice (16,720 J) is greater than Qsteam (0 J), this means that the 100 g of 100°C steam will be able to completely melt the 800 g of 0°C ice.
To show that 100 g of 100°C steam will completely melt 800 g of 0°C ice, we need to calculate the heat required to do so.
The process of melting ice requires heat transfer, which can be calculated using the formula:
Q = m * L
Where:
Q is the heat transfer (in joules)
m is the mass of the substance (in grams)
L is the specific latent heat (in J/g)
For ice, the specific latent heat of fusion (L) is approximately 334 J/g, which means that 334 joules of heat are required to change 1 gram of ice at its melting point into water.
First, let's calculate the heat required to convert 800 g of ice into water:
Q_ice = m_ice * L_ice
= 800 g * 334 J/g
= 267,200 J
Now, since steam at 100°C needs to be converted to water at 100°C before it can start melting the ice, we need to consider the heat required for this process.
The specific heat capacity of steam (C) is approximately 2.03 J/g°C. So, to convert 100 g of steam at 100°C to water at 100°C, we can use the formula:
Q_steam = m_steam * C_steam * (T_final - T_initial)
= 100 g * 2.03 J/g°C * (100°C - 100°C)
= 0 J
As we can see, no heat transfer is required to convert the steam into water at the same temperature.
Now, the total heat required to completely melt 800 g of ice is given by the sum of the two heat transfers:
Q_total = Q_ice + Q_steam
= 267,200 J + 0 J
= 267,200 J
Therefore, 100 g of 100°C steam will completely melt 800 g of 0°C ice by transferring a total of 267,200 joules of heat.