## Test 2 Question 7

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Amy Dinh 1A
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

### Test 2 Question 7

Can someone explain why cooling a hot cup of coffee after it was brewed spontaneous (delta G < 0) when melting ice is also spontaneous?

skyeblee2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

### Re: Test 2 Question 7

In both instances, they are approaching equilibrium with their surrounding temperature. It would require no input of energy for a cup of hot coffee to cool down (when it is hotter than its outside environment) or an ice cube to heat up (when it is cooler than its outside environment). Thus, both are spontaneous.

riddhiduggal
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

### Re: Test 2 Question 7

They are both spontaneous because they are both approaching equilibrium.

Anish Natarajan 4G
Posts: 34
Joined: Thu Jul 25, 2019 12:16 am

### Re: Test 2 Question 7

a cup of coffee is an open system in that the external temperature of the surroundings can affect the internal temperature of a system. A hot cup of coffee that exists in an area where the surrounding temperature is colder will allow for the heat transfer from the system to the surroundings.