dry steam at 100°c is bubbled into 500g of water originally at 20°c. what will be the temperature of the water after 30g of steam has condensed
specific latent heat of steam =2.26×10^6jkg–¹
specific heat capacity of water = 4200jkg–¹?
First, let's calculate the heat transferred from the steam to the water when 30g of steam condenses.
The heat transferred can be calculated using the formula Q = ml, where Q is the heat transferred, m is the mass of the steam condensed, and l is the specific latent heat of steam.
Q = 0.03kg * 2.26 x 10^6 Jkg^-1 = 67800 J
Next, let's calculate the final temperature of the water. We will assume no heat is lost to the surroundings.
The heat gained by the water can be calculated using the formula Q = mcΔT, where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Q = 500g * 4200 Jkg^-1°C^-1 * ΔT
67800 J = 500g * 4200 Jkg^-1°C^-1 * ΔT
ΔT = 67800 J / (500g * 4200 Jkg^-1°C^-1)
ΔT = 0.0323°C
Therefore, the final temperature of the water will be approximately 20.0323°C.