Newton's Law of Cooling of object

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I don't understand Newtons law of cooling

dT/dt = k(a - b)
were a is the object and b is the surroundings that the object is placed in...
Is this correct or is there suppose to be a negative sign like so
dT/dt = -k(a - b)

also does Newton's law of cooling apply to putting a cold object in a warmer enviorment were it is actually getting warmer and not cooling down? If so is there a negative sign or no? This negative sign is bothering me and I'm unsure what to do for when the object is placed in a warmer enviorment and is actually heating up and not cooling down

thanks for any help
 
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Newtons law of cooling states that 'rate of loss of HEAT ENERGY is proportional to the excess temperature'
This sounds logical and is in line with many other laws in physics where a rate is proportional to a difference of some sort
so dQ/dT = k(θ1-θ2)
 
ok so there is no negative sign regardless if it's cooling or warming up

and it's the object minus its surroundings?
 
Cooling implies that the object is hotter than the surroundings and the - sign indicates that the rate of heat loss will DECREASE as the temp difference decreases.
A + would indicate that the rate of heat loss (transfer!) would increase as the temp difference increased... sounds logical.
I am fairly certain someone will have a different view !
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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