Chemical Thermodynamics Question 426
Question: What is the amount of heat (in Joules) absorbed by 18 g of water initially at room temperature heated to $ 100{}^\circ C $ - If 10 g of Cu is added to this water, than decrease in temperature (in Kelvin) of water was found to be- C (p,m) for water $ 75.32J/molK $ ; C (p,m) for $ Cu=24.47J/molK. $
Options:
A) 5649,369
B) 5544,324
C) 5278,342
D) 3425,425
Show Answer
Answer:
Correct Answer: A
Solution:
- 18gm of water at $ 100{}^\circ C $
$ 10gm $ of Cu at $ 25{}^\circ C $ is added. $ q_{p}={C_{p,m}}dt $
$ =75.32\times \frac{J}{K,mol}\times \frac{18g}{18g/mol}( 373-298 )K $
$ =75.32\frac{J}{K}\times 75K $
$ =5.649\times 10^{3}J $ If now $ 10g $ of copper is added $ {C_{p,m}}=24.47J/mol,K $ Amount of heat gained by Cu $ =24.47\times \frac{J}{K,mol}\times \frac{10g}{63g/mol}( 373-298 )K $
$ =291.3J $ Heat lost by water $ =291.30J $
$ -291.30J=75.32\frac{J}{K}\times (T_2-373K) $
$ \Rightarrow -3.947K=T_2-373K $
$ \Rightarrow T_2=369.05K $
