Answer:
232 J/K
Explanation:
The amount of heat gained by the air = the amount of heat lost by the tea.
q_air = -q_tea
q = -mCΔT
q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)
q = 68,000 J
The change in entropy is:
dS = dQ/T
Since the room temperature is constant (isothermal):
ΔS = ΔQ/T
Plug in values (remember to use absolute temperature):
ΔS = (68,000 J) / (293 K)
ΔS = 232 J/K
Opposite charges attract
Like charges repel
The particle has 
and 
, and is undergoing a constant acceleration of 
.
This means its position at time 
is given by the vector function,
![\implies\vec r(t)=\left[4\,\mathrm m+\left(2\dfrac{\rm m}{\rm s}\right)t-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cimplies%5Cvec%20r%28t%29%3D%5Cleft%5B4%5C%2C%5Cmathrm%20m%2B%5Cleft%282%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29t-%5Cleft%281%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5Cright%5D%5C%2C%5Cvec%5Cimath-%5Cleft%281%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5C%2C%5Cvec%5Cjmath)
The particle crosses the x-axis when the 
component is 0 for some time 
, so we solve:
The negative square root introduces a negative solution that we throw out, leaving us with 
or about 3.24 seconds after it starts moving.
Im not sure if this is physics or mathematics. but if 300 msec then per minute this will equal to 300 X 60 sec =18000 m per minute
