Question

A 40 g block of ice is cooled to -69°C and is then added to 590 g of water in an 80 g copper calorimeter at a temperature of 22°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g·°C = 2090 J/kg°C.

Answer 1

Answer:

$T = 16.7 Degree C$

Explanation:

Heat released by water and copper calorimeter if they reach to Zero degree celcius from initial temperature

$Q = ms\Delta T$

$Q = 590(4.86)(22 - 0) + 80(0.385)(22 - 0)$

Q = 63082.8 + 677.6 = 63760.4 J

now heat required to raise the temperature of ice to 0 degree Celcius

$Q_1 = ms\Delta T$

$Q_1 = 40(2.090)(0 - (-69))$

$Q_1 = 5768.4 J$

now heat required to melt the ice

$Q_2 = mL = 40 (335 J/g) = 13400 J$

total heat required to melt the ice will be

$Q = Q_1 + Q_2 = 19168.4$

now remaining heat after whole system is at zero degree celcius is

$Q_{net} = 63760.4 - 19168.4 = 44592 J$

now the final temperature of the system is given as

$Q_{net} = ms\Delta T$

$44592 = (590 + 40)(4.186)(T - 0) + 80(0.385)(T - 0)[tex]T = 16.7 Degree C$