Question

A cube of ice is taken from the freezer at -5.5 ∘C and placed in a 75-g aluminum calorimeter filled with 300 g of water at room temperature of 20.0 ∘C. The final situation is observed to be all water at 17.0 ∘C. The specific heat of ice is 2100 J/kg⋅C∘, the specific heat of aluminum is 900 J/kg⋅C∘, the specific heat of water is is 4186 J/kg⋅C∘, the heat of fusion of water is 333 kJ/Kg.

What was the mass of the ice cube?Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

Explanation:

Let mass of ice cube taken out be m kg .

ice will gain heat to raise its temperature from - 5.5° to 0° and then from 0° to 17° .

Total heat gained = m x 2.1 x 5.5 + m x 333 + m x 4.186 x 17

= (11.55 + 333 + 71.162 )m

= 415.712 m kJ

Heat lost by aluminium calorimeter

= .075 x .9 x 3  

= .2025 kJ

Heat lost by water

= .3 x 4.186 x 3

= 3.7674

Total heat lost

= 3.9699 kJ

Heat lost = heat gained

415.712 m = 3.9699

m = .0095 kg

9.5 gm .

Answer 2

Answer:

0.00954g or 9.5x$10^{-3}$ kg

Explanation:

The only conversion that is needed is changing the heat of fusion of water from 333 kJ/kg to 333000 J/kg.

This is the condensed version of the equation needed for this problem: mcΔT + mL + mcΔT = mcΔT + mcΔT

This is the expanded version of the equation needed for this problem:

$m_{ice}$($c_{ice}$)(temperature of ice from -5.5°C to 0°C) + $m_{ice}$(L) + $m_{ice}$($c_{water}$)(temperature of water from 0°C to 17°C) = $m_{water}$($c_{water}$)(ΔT) + $m_{aluminum}$($c_{aluminum}$)(ΔT)

Use the equation to solve for the mass of ice:

m(2100)(5.5) + m(333000) + m(4186)(17) = 0.3(4186)(20-17) + 0.075(900)(20-17)

m [(2100x5.5) + 333000 + (4186x17)] = 3767.4 + 202.5

m(415712) = 3969.9

m = 0.00954g or 9.5x$10^{-3}$ kg