A 20.0 g piece of iron and a 20.0 g piece of gold at 100.0 ? C were dropped into
ID: 791419 • Letter: A
Question
A 20.0 g piece of iron and a 20.0 g piece of gold at 100.0? C were dropped into 1.0 L of water at 20.0? C. The molar heat capacitates of iron and gold are 24.19 J/ (mol ? ?C) and 25.41 / (mol ? ?C), respectively. What is the final temperature of the water and pieces of metal?
A 20.0 g piece of iron and a 20.0 g piece of gold at 100.0? C were dropped into 1.0 L of water at 20.0? C. The molar heat capacitates of iron and gold are 24.19 J/ (mol ? ?C) and 25.41 / (mol ? ?C), respectively. What is the final temperature of the water and pieces of metal?Explanation / Answer
Let T be the final temperature
Heat lost by metals = heat gained by water
Moles x molar heat capacity x temperature change of iron
+ moles x molar heat capacity x temperature change of gold
= mass x specific heat x temperature change of water
20/55.845 x 24.19 x (100 - T) + 20/196.97 x 25.41 x (100 - T)
= 1000 x 4.184 x (T - 20.0)
1124.335 - 11.243T = 4184T - 83680
4195.243T = 84804.335
Final temperature T = 20.21 deg C = 20.2 deg C
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