what is the final temperature (in °c) of a 94 g sample of iron with an initial temperature of 19.8 °c that…

Explanation:

Step1: Convert mass to kg

$m = 94g=0.094kg$

Step2: Use the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $Q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We know $Q=- 1068J$ (negative because heat is lost), $m = 0.094kg$, and $c = 440J/kg\cdot K$. Rearranging the formula for $\Delta T$ gives $\Delta T=\frac{Q}{mc}$. $\Delta T=\frac{-1068J}{0.094kg\times440J/kg\cdot K}\approx - 26K$. Since a change in Kelvin is the same as a change in Celsius, $\Delta T=-26^{\circ}C$.

Step3: Calculate the final temperature

The formula for $\Delta T=T_f - T_i$. Rearranging for $T_f$ gives $T_f=T_i+\Delta T$. Given $T_i = 19.8^{\circ}C$ and $\Delta T=-26^{\circ}C$. $T_f=19.8^{\circ}C+( - 26^{\circ}C)=-6.2^{\circ}C$

Answer:

$-6.2^{\circ}C$