I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.

THE FINAL TEMPERATURE OF WATER IS -4.117 °C

Explanation:

Mass of the aluminium = 50 g

c = 0.88 J/g C

Initial temperature of aluminium = 225 °C

Volume of water = 100 ml

Density of water = 1 g/ml

Mass of water = density * volume of water

Mass of water = 1 * 100 = 100 g of water

Initial temperature of water = 20 C

It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,

So therefore,

heat lost by aluminium = heat gained by water

H = mass * specific heat capacity * temeprature change

So:

m c ( T2- T1) = m c (T2-T1)

50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)

44 ( T2 - 225 ) = 418 ( T2 - 20)

44 T2 - 9900 = 418 T2 - 8360

-9900 + 8360 = 418 T2 - 44 T2

-1540 = 374 T2

T2 = - 4.117

So therefore the final temperature of water is -4.117 °C