Ask Question
4 January, 13:49

A 276.0 g piece of granite, heated to 596.0°C in a campfire, is dropped into 1.45 L water (d = 1.00 g/mL) at 25.0°C.

The molar heat capacity of water is cp, water = 75.3 J / (mol ·°C), and the specific heat of granite is cs, granite = 0.790 J / (g ·°C).

Calculate the final temperature of the granite in Celsius.

+4
Answers (1)
  1. 4 January, 17:44
    0
    Answer;

    Final temperature of the granite ≈ 44.82 °C

    Explanation;

    Let T be the equilibrium temperature (in °C) to be found.

    Moles of water;

    (1.45 L H2O) x (1000 mL/L) x (1.00 g/mL) / (18.01 g H2O/mol)

    = 80.512 mol H2O

    Heat gained by water;

    = (75.3 J / (mol ·°C)) x (80.512 mol) x (T - 25.0) °C

    = (6062.465 T - 151561.6)

    Heat lost by granite;

    = (0.790 J / (g ·°C)) x (276 g) x (596 - T) °C

    = (129951.84 - 218.04 T)

    Set the two expressions for heat gained/lost equal to each other:

    6062.465 T - 151561.6 = 129951.84 - 218.04 T

    6280.505 T = 281513.44

    T = 44.8234

    T ≈ 44.82 °C
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 276.0 g piece of granite, heated to 596.0°C in a campfire, is dropped into 1.45 L water (d = 1.00 g/mL) at 25.0°C. The molar heat ...” 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