Answer
$0.52 \frac{J}{g*K}$
Work Step by Step
Given that: $m_{Ti}= 20.8g$, $\Delta T_{Ti} = 24.3-99.5 = -75.2^{\circ}C$, $m_{H_{2}O} = 75.0g$, $\Delta T_{H_{2}O} = 24.3-21.7 = 2.6^{\circ}C$ $c_{H_{2}O} = 4.184 \frac{J}{g*^{\circ}C}$
Using Equation: $Q_{Ti} + Q_{H_{2}O} = 0$ we can calculate the $c_{Ti}$
$Q_{Ti} + Q_{H_{2}O} = 0$
$(m_{Ti}\times c_{Ti}\times\Delta T_{Ti})+(m_{H_{2}O}\times c_{H_{2}O}\times\Delta T_{H_{2}O})=0$
$(20.8g\times c_{Ti}\times -75.2^{\circ}C)+(75.0g\times4.184 \frac{J}{g*^{\circ}C}\times2.6^{\circ}C)=0$
$c_{Ti} = 0.52 \frac{J}{g*K}$
*Note because the formula requires the change in temperature the conversion of Celsius to Kelvin is not required
