Question

Q1. Lemonade can be cooled by adding lumps of ice to it. A student discovers that 70g of ice at a temperature of 0C cools 0.30kg of lemonade from 28C to 7C. The latent heat of fusion of ice is
0.33MJ/kg. The specific heat capacity of water is 4.2kJ/(kg k).
Determine
(a)the energy gained by the ice in melting,
(b) the energy gained by the melted ice,
(c) the energy lost by the lemonade,
(d) a value for the specific heat capacity of the lemonade.

Answer 1

Answer:

a) is 23.1 kJ, b) is 2.058 kJ, c) is 25158 J and d) is 3999 J/(kgK)

Explanation:

a) E=mLf

    = 70/1000 kg * 0.33MJ/kg

    =0.0231 MJ = 23.1 kJ

b) melted ice is water, so

E=mc(T1-T2)

=70/1000 kg * 4.2KJ/(kgK) * (28C-7C)                   (Kelvin and Celsius are on the same scale)

=2.058kJ

c)23.1kJ+2.058kJ (energy lost by lemonade= energy gained by water and ice)

=25.158kJ=25158J

d) E=mc(T2-T1)

c= E/m(T2-T1)

=25158J/(0.30kg*(28C-7C))

= 3999 J/(kgC) or 3999 J/(kgK)  as Kelvin and Celsius are on the same scale