## Essential University Physics: Volume 1 (3rd Edition)

$-10^{\circ}C$ $263.15 K$
We use the following equation to find: $\frac{h}{h_3} = \frac{T}{T_3}$ $\frac{57.8}{60} = \frac{T}{273.16}$ $T = -10^{\circ}C$ This is the same as 263.15 degrees Kelvin.