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

We find: $\frac{dV}{V} =\beta dT$ $lnV = \int_{T_1}^{T_2}( aT + .5BT^2 + \frac{c}{3}T^3)dT$ $lnV = \int_{0}^{12}( aT + .5BT^2 + \frac{c}{3}T^3)dT$ $V=1.00L$ Thus, we find: $ln \frac{V_2}{V} = 12a + 77b + 576c$ Plugging in the values of a, b, c, and V (which was found above), we obtain: $V_2=1.00L$