- #1
tuki
- 19
- 1
- Homework Statement
- 1.00 kg of hot water (T = 100 celcius) and 1.00 kg of cold water (T = 0 celcius) are mixed in bowl. It results in 2.00kg of water which temperature is 50 Celcius. Calculate the whole systems change in entropy.
- Relevant Equations
- Change in entropy:
$$ dS = \frac{dQ}{T} $$
Heat capacity:
$$ Q = Cm\Delta T $$
I tried following:
$$ dS_{\text{total}} = |\frac{dQ}{T_c}| |\frac{dQ}{T_H}| $$
where ## T_h ## is temperature of hot water and ## T_c ## is temperature of cold water. Coefficient for water wasn't provided in the assignment so i used following value c = 4190 J/kgK.
$$ dS_{\text{total}} = |\frac{cm\Delta T}{T_c}| - |\frac{cm\Delta T}{T_h}| $$
where ## dQ = cm\Delta T = 4190 \text{ J/kgK} \cdot 1 \text{ kg} \cdot 50 \text{ K} ##
my result for change in entropy is:
$$ dS \approx 205 \text{ J/K} $$
However the correct answer should be:
$$ dS \approx 102 \text{ J/K} $$
$$ dS_{\text{total}} = |\frac{dQ}{T_c}| |\frac{dQ}{T_H}| $$
where ## T_h ## is temperature of hot water and ## T_c ## is temperature of cold water. Coefficient for water wasn't provided in the assignment so i used following value c = 4190 J/kgK.
$$ dS_{\text{total}} = |\frac{cm\Delta T}{T_c}| - |\frac{cm\Delta T}{T_h}| $$
where ## dQ = cm\Delta T = 4190 \text{ J/kgK} \cdot 1 \text{ kg} \cdot 50 \text{ K} ##
my result for change in entropy is:
$$ dS \approx 205 \text{ J/K} $$
However the correct answer should be:
$$ dS \approx 102 \text{ J/K} $$