Question

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Constants Periodic Table Part A An insulated beaker with negligible mass contains liquid water with a mass of 0.250 kg and a temperature of 60.1 C How much ice at a temperature of-16.8 must be dropped into the water so that the final temperature of the system will be 24.0 C Take the specific heat of liquid water to be 4190 /kg , the specific heat of ice to be 2100 Jkg-K and the heat offusion for water to be 3.34x105 J/kg View Available Hint(s) mice. 10.098 kg Submit Previous Answers Incorrect; Try Again; 5 attempts remaining

Answer 1

Here ,

let the mass of ice needed is m

Now, heat lost by water = heat gain by ice

0.250 * 4190 * (60.1 - 24) = m * (2100 * 16.8 + 3.34 *10^5 + 4190 *
24)

solving for m

m = 0.0805 Kg

the mass of ice needed is 0.0805 Kg