Question

If 20g steam initially at 100°c is added to 60g of ice initially at 0°c , then find the equilibrium temperature of the mixture.

Answer 1

Dear Student,

◆ Answer -

T = 25 ℃

● Explanation -

# Given -

T1 = 100 ℃ = 373 K

T2 = 0 ℃ = 273 K

m1 = 20 g

m2 = 60 g

# Solution -

Final equilibrium temperature of mixture is -

T = (m1T1 + m2T2) / (m1 + m2)

T = (20×373 + 60×273) / (20 + 60)

T = 23840 / 80

T = 298 K

T = 25 ℃

Hence, final equilibrium temperature of mixture is 25 ℃.

Thanks dear...

Answer 2

Answer:

◆ Answer -

T = 25 ℃

● Explanation -

# Given -

T1 = 100 ℃ = 373 K

T2 = 0 ℃ = 273 K

m1 = 20 g

m2 = 60 g

# Solution -

Final equilibrium temperature of mixture is -

T = (m1T1 + m2T2) / (m1 + m2)

T = (20×373 + 60×273) / (20 + 60)

T = 23840 / 80

T = 298 K

T = 25 ℃

Hence, final equilibrium temperature of mixture is 25 ℃.