## College Physics (4th Edition)

We can use this formula to convert from $^{\circ}J$ to $^{\circ}C$: $C = \frac{192}{144}J - 114$
The temperature range from the freezing point to the boiling point is $0^{\circ}J$ to $144^{\circ}J$, which corresponds with $-114^{\circ}C$ to $78^{\circ}C$. The temperature change from the freezing point to the boiling point is $+144^{\circ}J$ which is $+192^{\circ}C$ To convert from $^{\circ}J$ to $^{\circ}C$, we need to multiply by a factor of $\frac{192}{144}$. Since $0^{\circ}J$ corresponds with $-114^{\circ}C$, we also need to subtract a value of $114$ We can use this formula to convert from $^{\circ}J$ to $^{\circ}C$: $C = \frac{192}{144}J - 114$ We can check this formula with the freezing point and boiling point: $C = \frac{192}{144}(0) - 114 = -114^{\circ}C$ $C = \frac{192}{144}(144) - 114 = 78^{\circ}C$