Question

If steam of mass 100 g and temperature is 100 degree celcius is released on the ice slab of temperature 0 degree celcius how much ice will melt​

Answer 1

Answer:

mass of ice m=?,L2= 80 cal/g ,c (water ) = 1cal/g∘C

According to the principle of heat lost by hot body = heat gained by cold body .

Conversion of steam into water :

Q1=m1L91)=100g×540cal/g=54000cal

Decrease in the temperature of this water to 0∘C

Q2=m1c×(T1−0∘C)=100g×1cal/g∘C×(100∘C−0∘C)=1000cal

Melting of ice = Q3=mL2=m×80cal/g

Now , Q1+Q2=Q3

( 54000 + 10000) cal. = m×80 cal/g

m=6400080g=800g

800 g of ice will melt.