Answer:
[tex]8.4 \times 10^{5}\; \rm J[/tex], assuming that there's no heat exchange between the washing machine and the environment.
Explanation:
Let [tex]m[/tex] denote the mass of water and [tex]c[/tex] the specific heat capacity of water. The energy required to raise the temperature of that much water by [tex]\Delta T[/tex] would be:
[tex]Q = c \cdot m \cdot \Delta T[/tex].
Washing at [tex]30\; \rm ^{\circ} C[/tex] would require a temperature change of [tex]\Delta T = 30\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 14\; \rm K[/tex].
Washing at [tex]50\; \rm ^{\circ} C[/tex] would require a temperature change of [tex]\Delta T = 50\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 34\; \rm K[/tex].
In both situations, [tex]c = 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1}[/tex] while [tex]m = 10\; \rm kg[/tex].
Calculate the energy required in either situation:
Washing at [tex]30\; \rm ^{\circ} C[/tex]:
[tex]\begin{aligned}& Q({30\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 14\; \rm K \\ &= 588000 \times 10^{5}\; \rm J\end{aligned}[/tex].
Washing at [tex]50\; ^{\circ} {\rm C}[/tex]:
[tex]\begin{aligned}& Q({50\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 34\; \rm K \\ &= 1428000 \; \rm J\end{aligned}[/tex].
[tex]1428000\; \rm J - 588000\; \rm J = 8.4 \times 10^{5}\; \rm J[/tex].
