PHY..............

Question

The mass of hot soup is 0.3 kg . The specific heat capacity and temperature of the soup are 3600 J kg-1'C-1 and 90'C respectively . How much water at 2'C should be added to lower the temperature of the soup to 30'C ?

Answer 1

soup:
E = mcDT
E = 0.3(3,600)(30-90)
= -64,800J

let specific heat capacity of water is 4200J kg-1'C-1

water:
m = E / (cDT)
= -64,800/ (4200)(90-2)
= 0.1753kg

Answer 2

答錯了的是最佳回答，但答對了的卻是其他回答。

水的溫度是由2度升30度，唔係90度呀！！！

Answer 3

Let m kg be the mass of water required.

The soup changes from 90oC to 30oC; The water changes from 2oC to 30oC.
Energy released by soup = 0.3 x 3600 x (90-30) = 64800 J
Energy absorbed by water = m x 4200 x (30-2) = 117600m J

Energy absorbed by water = Energy released by soup
117600m = 64800
Hence, m = 0.551
The mass of water required = 0.551 kg