## College Physics (4th Edition)

(a) The temperature change in kelvins is $-6.0~K$ (b) The temperature change in $^{\circ}F$ is $-10.8~^{\circ}F$
(a) Let the two temperatures be $C$ and $C-6.0$. We can convert the two temperatures to kelvins: $K_1 = C+273.15$ $K_2 = (C-6.0)+273.15 = C+267.15$ We can find the temperature change in kelvins: $\Delta T = K_2-K_1$ $\Delta T = (C+267.15)- (C+273.15)$ $\Delta T = -6.0~K$ The temperature change in kelvins is $-6.0~K$ (b) Let the two temperatures be $C$ and $C-6.0$. We can convert the two temperatures to $^{\circ}F$: $F_1 = \frac{9}{5}C+32$ $F_2 = \frac{9}{5}(C-6.0)+32 = \frac{9}{5}C+21.2$ We can find the temperature change in $^{\circ}F$: $\Delta T = F_2-F_1$ $\Delta T = (\frac{9}{5}C+21.2)- (\frac{9}{5}C+32)$ $\Delta T = -10.8~^{\circ}F$ The temperature change in $^{\circ}F$ is $-10.8~^{\circ}F$