Question

190 g of ice at 0 ºC is mixed with 49 g of steam at 100 ºC to make water. The latent heat of fusion of water is 33.5 × 104 J/kg, and the specific heat of water is 4186 J/kg∙K, the latent heat of vaporization of water is 22.6 × 105 J/kg.

A) Determine the amount of heat released by steam at 100*C to make water at 100*C.

B) Determine the amount of heat absorbed by ice at 0*C to make water at 0*C.

C) What is the final temperature of the water?

Answer 1

here,

the mass of ice ,m1 = 190 g

the mass of steam , m2 = 49 g

A)

the amount of heat released by steam at 100 degree C to make water at 100 degree C , Qa = m2 * Lv

Qa = 49 * 2256 J = 1.1 * 10^5 J

B)

the amount of heat absorbed by ice at 0 degree C to make water at 0 degree C , Qb = m1 * Lf

Qb = 190 * 335 J = 6.37 * 10^4 J

C)

let the final temperature be Tf

using principle of calorimetry

Qb + m1 * Cw * ( Tf - 0 ) = Qa + m2 * Cw * ( 100 - Tf)

6.37 * 10^4 + 190 * 4.186 * ( Tf - 0 ) = 1.1 * 10^5 + 49 * 4.186 * ( 100 - Tf)

solving for Tf

Tf = 66.78 degree C

the final equilibrium temperature is 66.78 degree C