Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.345 kg and a temperature of 64.3 ∘C .

Part A

How much ice at a temperature of -24.7 ∘C must be dropped into the water so that the final temperature of the system will be 21.0 ∘C ?

Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×10⁵ J/kg .

Answer 1

Let the unkown mass of ice is M,

Q = m*specific heat*delta(T)

Amount of heat lost by the 0.300 kg of water in the beaker to the ice,

Q = 0.345*4190*(64.3 - 21) = 62592.3 J

Amount of heat gained by the ice will be,

Q(ice) + Q(fusion) + Q(liquid) for the unknown mass M

where,

Q(ice) = M*2100*(0 - (-24.7)) = M*2100*24.7 = 51870M

Q(fusion) = M*3.34*10^5 = 3.34*10^5M

Q(liquid) = M*4190*(21 - 0 ) = 87990M

When equilibrium is reached,

62592.3 J = 51870M + 3.34*10^5M + 87990M = 473860 M

M = 62592.3 /473860 = 0.132 kg