Two 20.0-g ice cubes at –13.0 °C are placed into 285 g of water at 25.0 °C. Assu

Question

Two 20.0-g ice cubes at –13.0 °C are placed into 285 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Two 20.0-g ice cubes at-130°C are placed into 285 g of water at 250°C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts. heat capacity of H20(s) 37.7J/(mol K) heat capacity of H20(0) 75.3 J(mol K) enthalpy of fusion of H20 6.01 kJ/mol Number

Explanation / Answer

Given that

mass of Ice = 2 x 20 g = 40.0 g

mass of water = 285 g

Given that Two 20.0-g ice cubes at –13.0 °C are placed into 285 g of water at 25.0 °C.

Therefore,

Heat lost by water at 25 °C = Heat gained by ice at -13 °C

Heat loss or gain Q = mcdT , c= specific heat

specific heat of water = 4.18 J/g/oC

specific heat of ice = 2.03 J/g/oC

Let T be the final temperature of the water

Hence,

285 g x 4.18 J/g/oC x [ 25 - T ] = 40 g x 2.03 J/g/oC x [T-(-13)]

29782.5 - 1191.3 T = 81.2 T + 1055.6

1272.5 T = 28726.9

T = 22.57 oC

Therefore, final temperature of the water =   22.57 oC

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