Randall D. Knight — Work, Heat, and the First Law of Thermodynamics: Problem 84

For an object that cools by convection, the temperature \( T \) often can be described by the equation \[ \frac{dT}{dt} = -b(T - T_{\text{env}}) \] where \( T_{\text{env}} \) is the temperature of the surrounding environment and \( b \) is an experimentally determined constant. The left side, \( dT/dt \), is the rate at which the temperature changes. The equation says that the rate of change is proportional to how much the object’s temperature differs from its environment: a rapid change when the temperature difference is large, slowing to a much smaller change when the temperature difference is small. This agrees with our everyday experience. The minus sign indicates that the object’s temperature falls, a negative rate of change, when it is hotter than its environment. This model of convection is often called Newton’s law of cooling. To solve the equation, change the variable to \( U = T - T_{\text{env}} \), for which \( dU = dT \). Then integrate from \( U_i = T_0 - T_{\text{env}} \) at \( t_i = 0 \) to \( U_f = T - T_{\text{env}} \) at a later time \( t_f = t \), where \( T_0 \) is the object’s initial temperature at \( t = 0 \). The result is an expression for the temperature difference \( T - T_{\text{env}} \) as a function of time. Suppose a mug of hot coffee in a \( 20^{\circ}\text{C} \) room cools from \( 80^{\circ}\text{C} \) to \( 60^{\circ}\text{C} \) in \( 5.0\text{ min} \). How long will it take to cool an additional \( 20^{\circ}\text{C} \), from \( 60^{\circ}\text{C} \) to \( 40^{\circ}\text{C} \)?