Whiskey Stones are an alternative method to ice cubes for chilling ones drink wi

Question

Whiskey Stones are an alternative method to ice cubes for chilling ones drink with less dilution. Whiskey stones cubes of soapstone that can be kept in the freezer right along with your ice trays. Determine the final temperature and the change in enthalpy if a: 3oz glass of whiskey initially at room temperature (20 degree C) was to be chilled by four 8 cm3 whiskey stone at -15 degree C 5oz glass of whiskey initially at room temperature (23 degree C) was to be chilled by 32mL of ice cubes at -15 degree C. *BONUS 3oz glass of whiskey initially at room temperature (20 degree C) was to be chilled by 32mL of ice cubes at -15 degree C. Ignore heat lost to the glass and the environment. Note: Whiskey: density: .9227 kg/L, heat capacity: 3.4 1/g degree C Ice: density: .9167 g/cm3, heat capacity(ice): 2.11 1/g degree C' heat capacity (water): 4.18 1/g degree C' Latent heat of fusion: 334 1/g Stones: density 2.980 g/cm3' heat capacity: 0.98 1/g degree C'

Explanation / Answer

a) 3oz = 85.049g of whiskey

inital temp = 20c

8cm^3 whisky stone = 8*2.98 =23.84 g of whiskey stones

inital temp = -15 c

let the final temp be t

heat lost by whiskey = heat gained by stones

85.049*3.4*(20-t) = 23.84*0.98*(t+15)

5783.332-289.1666t=23.3632t+350.448.

312.5298t=5432.884

t=17.3835 celsius

b)

5oz = 141.75g

initial temp = 23 celsius

32 ml ice cube=32*0.9167=29.3344 g of ice

initial temp = -15c

let the final temp be t

heat lost by whiskey = heat gained by stones

141.75*3.4*(23-t) = 29.3344*2.11*(0+15) +29.3344*344 + 29.3344*4.18*(t)

11084.85-481.95t=928.43376+10091.0336+122.617792t

604.567792t=65.38264

t=0.10814 celsius

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