Answer:
Explanation: A
[tex]$The heat $Q$ which a body absorbs or radiates can be expressed as:$Q=m c \Delta T$Where $m$ is the mass, $c$ is the specific heat capacity and $\Delta T$ is the change in temperature.[/tex]
[tex]$Substitute the known and given values into the equation above and calculate the results:[/tex]
[tex]\begin{aligned}Q_{w} &=m_{w} c_{w} \Delta T=200 \cdot 1000 \cdot(20-10)=2 \cdot 10^{6} \mathrm{~J} \\Q_{A l} &=m_{A l} c_{A l} \Delta T=200 \cdot 897 \cdot(20-10)=1.8 \cdot 10^{6} \mathrm{~J} \\Q_{s} &=m_{s} c_{s} \Delta T=100 \cdot 420 \cdot(20-10)=4.2 \cdot 10^{5} \mathrm{~J}\end{aligned}[/tex]
[tex]\text { Therefore, } 200 \mathrm{~kg} \text { of water would require the most energy to heat up. }[/tex]
