Pls help pls help me solve. Dry steam at 100 degree Celsius is bubbled into 500 g of water originally at 20 degree Celsius. What will be the temperature of the water after 30 g of steam has condensed? ( specific heat capacity of water =4200,specific latent heat of steam =2260000
sum of heats absorbed is zero.
30*2260000 +30(4200)(tf-20)-500(tf-20)=0
Pls solve
bob...how about the 100degrees celcius
I no know abeg.
To solve this problem, we need to consider the heat exchange between the steam and the water. Let's break it down into steps:
Step 1: Calculate the heat transferred when the steam condenses.
We know that 30 g of steam has condensed. To calculate the heat transferred during condensation, we'll use the specific latent heat of steam, which is given as 2,260,000 J/kg.
The heat transferred (Q) can be calculated using the formula:
Q = mass × specific latent heat
First, we need to convert the mass of steam from grams to kilograms:
mass of steam = 30 g = 0.03 kg
Now, we can calculate the heat transferred:
Q = 0.03 kg × 2,260,000 J/kg = 67,800 J
Step 2: Calculate the heat gained by the water.
The heat gained by the water can be calculated using the formula:
Q = mass × specific heat capacity × change in temperature
The initial mass of water is given as 500 g, which is equal to 0.5 kg.
The specific heat capacity of water is given as 4200 J/kg·°C.
The change in temperature is the final temperature minus the initial temperature.
Let's assume the final temperature is T °C. The initial temperature is given as 20 °C.
Therefore, we can write the equation as:
67,800 J = 0.5 kg × 4200 J/kg·°C × (T - 20 °C)
Now, we can solve for T:
67,800 J = 2100 kg·°C × (T - 20 °C)
67,800 J = 2100 kg·°C × T - 42,000 J
2100 T = 109,800 J
T = 109,800 J / 2100 kg·°C
T ≈ 52.29 °C
Therefore, after 30 g of steam has condensed, the temperature of the water would be approximately 52.29 °C.