Question

ice at 0.0 degree C is used to cool water. What is the minimum mass of ice required to cool 325 g of water from 30.5 degree C to 4.0 degree C? (Heat of fusion= 333 J/g; specific heat capacities: ice= 2.06 J/g.K, liquid water= 4.184 J/g.K)

a. 108g b. 125g c. 325g d. 605g e. 1.75x10^4

Show me the full work.

Answer 1

Amount of heat relesed when the water is dropped from 30.54^oC to 4^oC

Q = - m_water C_water dT = - 325 g x 4.184 J/g.K x 26.5 K = - 36034.7 J     

36034.7 J of energy is absorbed by unknown amount of ice at 0^oC

               = Latent heat of ice fusion of given weight + heat required to raise temp of melted ice from 0^oC to 4^oC

               = ( m_ice x L ) +   ( m_water x C_water x dT )

               = m [L + (Cwater x dT)]

36034.7 J = m [ 333 J/g + (4.184 J/g.K x 4K)]                     4K change in temperature of water in K

36034.7 J = m [ 349.736 J/g]

36034.7 J / 349.736 J/g = m

103.03 g   = m