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(Solved): show full work, answer fast. i will give good rating. A bowl of soup at \( 80^{\circ} \mathrm{C} \) ...



show full work, answer fast. i will give good rating.


A bowl of soup at \( 80^{\circ} \mathrm{C} \) sits in a \( 20^{\circ} \mathrm{C} \) room and cools to \( 74^{\circ} \mathrm{C
A bowl of soup at \( 80^{\circ} \mathrm{C} \) sits in a \( 20^{\circ} \mathrm{C} \) room and cools to \( 74^{\circ} \mathrm{C} \) after 1 minute. Using the model for Newton's Law of Cooling \( T(t)=\left(T_{0}-T_{s}\right) e^{k t}+T_{s} \), where \( T_{0} \) is the initial temperature of the object and \( T_{s} \) is the temperature of the surroundings, write the equation for the temperature of the bowl of soup after \( t \) minutes in the ropm. You will need to determine the value of \( k \) using the given information.


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Given that a bowl of soup is at 80*C sits in a 20*C room and cools to 74*C
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