Ask Question
21 March, 01:58

A 94.7-g sample of silver (s = 0.237 j / (g · °c)), initially at 348.25°c, is added to an insulated vessel containing 143.6 g of water (s = 4.18 j / (g · °c)), initially at 13.97°c. at equilibrium, the final temperature of the metal-water mixture is 22.63°c. how much heat was absorbed by the water? the heat capacity of the vessel is 0.244 kj/°c.

+1
Answers (1)
  1. 21 March, 04:22
    0
    Q = m*C*dT

    m = mass

    C = specific heat capacity

    dT = Temperature change

    Q = heat evolved

    Heat lost by metal is equal to the sum of heat taken up by water and vessel. Since water is taken in the vessel so

    the initial and final temperatures would be same for vessel and water ...

    Q = 94.7 * 0.237 * (22.63 - 348.25) = 7308.18 Joules = 7.3 KJ

    Heat lost by silver = 7.3 KJ

    Heat absorbed by water = 143.6*4.18 * (22.63 - 13.97) = 5.2 KJ
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 94.7-g sample of silver (s = 0.237 j / (g · °c)), initially at 348.25°c, is added to an insulated vessel containing 143.6 g of water (s = ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers