Question

When you mix two substances that are initially at different temperatures, the hotter substance cools and the colder substance warms until they reach a common temperature. Then the substances are said to be in thermal equilibrium. For our purposes assume that the heat is exchanged only between the hot and cold materials and that no heat is lost to the surroundings.

The specific heat capacity of water is 1.0 cal/(g Co) and the specific heat capacity of copper is 0.092 cal/(g Co) 1)What would be the final temperature when 110 g of hot water at 80 oC is combined with 120 g of cold water at 30 oC. Tfinal =

Answer 1

In the given question 110g of hot water at 80 c is combined with 120g of cold water at 30°c. In this situation the hot water cools and lost heat energy and the cold water warms and gaind heat energy and they reach a common final temperature Now, Heat energy lost by the hot water is 8, = mi asx TX LAT) (Ti-Tfinal) Where me in the mass of hot water. (m. = 110g) i sis the specific heat capacity of water. (3=1.0 cal/g c) To is the initial temperature of a hot water (T, = (80 +273)= 353k) 8, = 210 x 1 x Mind (353- Tfinal) 8. = 110 ( 353 - Tfinal) Heat energy gaind to by the cold water is : 2 = M2 xs x (Tfinal - T2) where ma is mass of cold water (m=1206) To is the initial temperature of the cold water. (T2 = 30 + 273 = 303k) 9₂ = 120 (Tfinal - 303) In thermal equilibrium S, and a, must be equal L. , 9, = 02 a 118 (353- Tfinal) = 126 (Tfinal 303) 3883 - 11 Tfinal = 12 Tpinal - 3636 23 Tfinal = 7519 Tfinal = 3276 - 540