Question

Review 1 Constants l Periodic Table Part A An insulated beaker with negligible mass contains liquid water with a mass of 0.200 kg and a temperature of 71.2 How much ice at a temperature of-17.0 °C must be dropped into the water so that the final temperature of the system will be 37.0 C? Take the specific heat of liquid water to be 4190 J/kg. K, the specific heat of ice to be 2100J/kg K, and the heat of fusion for water to be 3.34x105 J/kg View Available Hint(s)

Answer 1

Solution)

Let the Mass of the ice be x kg

Now, Heat lost by water, Q1 = mcdelta t

= 0.200 * 4190 * ( 71.2-37) = 28659.6 J

Now, Heat gained by ice, Q2 = m Ci Δt + m Li + m Cw Δt

= x [ 2100 * 17 + 334 x 10^3 + 4190 * 37 = x * 524730

We know,

From the principle of method of mixtures

x * 524730= 28659.6

Mass of the ice, x = 0.0546 kg  = 54.6 g (Ans)

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