Question SPM.2.HL.TZ0.8a

$\left(0.030\times 0.7\times {10}^3\times 5\right)+\left(0.030\times 220\times {10}^3\right)+\left(0.030\times 2.13\times {10}^3\right)\left(T-47\right)$

$=\left(0.150\times 2.13\times {10}^3\right)\left(240-T\right)$

One heat capacity term correctly substituted ✓

latent heat correctly substituted $\left(0.030\times 220\times {10}^3\right)$ ✓

$T=190$ «°C» ✓