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 $



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