How many joules of heat are lost by 3580 kg granite as it goes from 41.2°c to -12.9°c

The granite loses 1.53 × 10⁸ J of heat.

The formula for the heat (q) transferred is

q = mCΔT

I shall assume that the specific heat capacity of granite is 0.790 J·°C⁻¹g⁻¹.

m = 3.580 × 10⁶ g; ΔT = T_f – T_i = -12.9 °C - 41.2 °C = -54.1 °C

∴ q = 3.580 × 10⁶ g × 0.790 J·°C⁻¹g⁻¹ × (-54.1 °C) = -1.53 × 10⁸ J

The granite block loses 1.53 × 10⁸ J.

where are the choice options?