were you taught this formula or the differential equation?venger said:Homework Statement
"A cup of hot chocolate, in a room temperature of 21*C, cools accordiing to Newton's law of cooling. Determine the rate of cooling, k, of the hot chocolate if it cools from 86*C to 65*C in 15 minutes"Homework Equations
Newton's cooling law, ln function
T-Ts=(T. - Ts)e^(kt)The Attempt at a Solution
T-Ts=(T. - Ts)e^(kt)
65-21=(86-21)e^(15k)
44=65e^(15k)
44/65=e^(15k)
ln(44/65)=15k
(ln(44/65))/15=k
k=-0.02601...
Is this right?
Newton's Cooling Law is a mathematical equation that describes the rate at which a hot object cools in a cooler environment. It states that the rate of change of temperature of an object is proportional to the difference between its temperature and the temperature of its surroundings.
Newton's Cooling Law is used in a variety of fields, including physics, chemistry, and engineering. It can be used to study the cooling of hot liquids, the cooling of electronic components, and even the cooling of the Earth's surface.
The main factors that affect the rate of cooling are the initial temperature of the object, the temperature of its surroundings, and the thermal properties of the object, such as its specific heat capacity and thermal conductivity.
The formula for Newton's Cooling Law is dT/dt = -k(T-Ts), where dT/dt is the rate of change of temperature, k is a proportionality constant, T is the temperature of the object, and Ts is the temperature of its surroundings.
Newton's Cooling Law can be applied in various real-life situations, such as predicting the cooling of a cup of hot coffee, designing efficient cooling systems for electronic devices, and understanding the cooling of the Earth's atmosphere. It can also be used to determine the time it takes for food to cool down to a safe temperature to eat.