Question

A typical eruption temperature of basaltic lava is about 1200°C. Assuming that basaltic magma travels from its source region in the mantle quickly enough so that negligible heat is lost to wall rocks, calculate the temperature of the magma at a depth of 40 km. The density (δ) of basaltic magma at 1200°C is 2610 kg/m3 (remember that V=1/δ); the coefficient of thermal expansion (α) is about 10-4 /K. Assume a heat capacity (CP) of 850 J/kg-K and that pressure is related to depth as 1 km = 33 MPa (surface pressure is 0.1 MPa). (HINT: “negligible heat loss” means the system may be treated as adiabatic)(also, 1 MPa = 1x106 kg/m-s2, you will need this to convert pressures)
Adiabatic means constant entropy, so we can start with a substituted equation of:

$\oint_{1473}^{T_{1}} \frac{dT}{T} = \oint_{0.1}^{1320}\frac{\alpha V}{C_{P}}dP$

We can thus simplify this expression to the following:

$ln(T_{1}) = \frac{\alpha V}{C_{P}}* (P-P_{0}) + ln(1473)$

Don’t forget that you don’t have V, but you do have density (δ). Also, you will need to convert P to kg/m-s2 in order to obtain the appropriate units.

Answer 1

Given T_0 = 1473k h_1 = 40km p_1 = 40 times 33 times 10^6 kg/ms^2 p_0 = 0.1Mpa = 0.1 times 10^6 kg/ms^2 s = 2610 kg/m^3 so V = (1/2610) M^3/kg alpha = 10_4(K)^1 C_p = 850 J/kg-k from the given equation l_n T_1 = 10^-4 times (1/2610)/850 (40 times 33 - 0.1)10^6 + l_n(1473) l_n T_1 = 7.354 T_1 = e^7.354 = 1563.3k