# A cup of coffee, cooling off in a room at temperature 20 degrees celsius, has a cooling constant of k=0.09min^-1. Use linear approximation to estimate the change in temperature over the next 6 seconds when T=80 degrees celsius.

Newton's Law of cooling states that the temperature changes with a speed proportional to the difference of the temperatures. The corresponding formula is `T ( t ) = T_R + ( T_0 - T_R ) e^( - k t ) , ` where `T_R = 20^@ C ` is the...

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Newton's Law of cooling states that the temperature changes with a speed proportional to the difference of the temperatures. The corresponding formula is `T ( t ) = T_R + ( T_0 - T_R ) e^( - k t ) , ` where `T_R = 20^@ C ` is the room temperature, `T_0 = 80^@ C ` is the initial cup temperature, `k = 0.09 min^( -1 ) ` is the cooling constant, and `t ` is the time in minutes.

The formula itself gives the answer `T ( 6 ) = 20 + 60 e^( - 0.54 ) ` `approx 55^@ C .`

But the question is about linear approximating the formula, which is the same as to use the tangent line at `t = 0 . ` This tangent line has the equation `T_( lin ) ( t ) = T ( 0 ) + t * T ' ( 0 ) . `

Naturally, `T ( 0 ) = T_0 = 80^@ C . ` Also, `T ' ( t ) = -k ( T_0 - T_R ) e^( - k t ) = - 5.4 e^( - 0.09 t ) , ` thus `T ' ( 0 ) = - 5.4 ( (^@ C ) / min ) .`

This way, the linear approximation gives us `T ( 6 ) approx 80 - 5.4 * 6 = 47.6 (^@ C ) .`

It's not a surprise that the approximated temperature is lower: the linear approximation does not take into account the slowdown of cooling as the difference of temperatures decreases.

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