The overall heat capacity of the calorimeter is $8\mathrm{.}67k\frac{J}{}C.$ If the calorimeter contained $1\mathrm{.}500kg$ of water, what is the heat capacity of the dry portion of the calorimeter?

$C_{\text{c}\text{a}\text{l}(complete)}=C_{\text{c}\text{a}\text{l}(dry)}+C_{\text{c}\text{a}\text{l}(H_{2}O)}$

$C_{H_{2}O}=4\mathrm{.}184\frac{J}{g}C$

$8\mathrm{.}67k\frac{J}{}C=C_{\text{c}\text{a}\text{l}(dry)}+(1\mathrm{.}500(kg)4\mathrm{.}184\frac{kJ}{k}gC)$

$C_{\text{c}\text{a}\text{l}(dry)}=k\frac{J}{}C$

Enter the magnitude. Use significant figures.