Question

How much ice (in grams) would have to melt to lower the temperature of 354 mL of water from 25 ∘C to 6 ∘C? (Assume the density of water is 1.0 g/mL.)

Answer 1

Volume of water = 354 mL

density of water = 1.0 g/mL

Mass of water(m) = volume * density = 354 mL* 1.0 g/mL = 354 g

Change in temperature (T) = 25-6 = 19^oC

Specific heat of water(c) = 4.184 J/g.^oC

Heat released (q) = m*c* T = 354 g *4.184 J/g.^oC * -19^oC = -28141.58 J = -28.14 kJ

Heat released by water = Heat taken by ice

-q_water = q_ice = 28.14 kJ

Ice is initially at a temperature of 0^oC and then its temperature rises to 6^oC

So, q_ice = Latent heat of fusion + Heat required to increase the temperature from 0 to 6^oC

Heat of fusion = 334 J/g

q_ice = ( mass* 334) + (m*c* change in temperature)

q_ice = ( mass* 334) + (m*4.184 J/g.^oC *6^oC)

28141.58 J = 334 m + 25.104 m = 359.104 m

m = 78.4 g