Question

structure.com/courses/1253444/quizzes/2442703/take Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in a calorimeter. The initial temperature of the water is 21.2°C. To the calorimeter a 29.458 g piece of metal at 98.9 °C is added. The final temperature of the contents of the calorimeter is measured to be 29.5°C. (HINT: the specific heat of water is 4.184 ) heat absorbed by water = Knowing the amount of heat the water absorbs is equal and opposite to the heat lost by the metal, find the specific heat of the metal. Cmetal

Answer 1

Rise in the temperature of water = (t-t, jºc where t, is the initial temperature of water And t is the final temperature of the mixture, So, (+-t,) c = (29.5 - 21.2°C A n ula - 281.3K Now heat energys gained by water = me, (x-to) where my is the mass of water and C, is the specific heat of water So, heat absorbed by water = 15gx4 1845/gk x 281.3K - 88 271.95 = 88.27 kJ

m Now from the principle of calosimetry, heat absorbed by water = heat lost by the metal So, loss of temperature of the metal = (tzt) oc = (98-9- 29.5) ¢ S = 69.4°C = 342.4K Here ta is the initial temperature of the metal.). So, heat energy lost by the metal = mzcz (tzt) where me is the mass of the metal and a is the specific heat of the metal.

Usings the principle of calorimetry we can write, m₂ C₂ (tzt) = 88.27 kJ OX = 29:458 g x C₂ x 342.4K = 88127 kJ - 88.27 2.29.458x34217 kJ/gk. = 0.00 8751 kJ I gk = 8.751 lak Hence specific heat of the metal = 8.7515/gk. So, heat absorbed by water = 88.27 kJ Z (Aus) And Specific heat of the metal = 8.751 J/g kjer