a sample of an unknown substance has a mass of 89.5 g. if 345.2 j of heat are required to heat the substance…

a sample of an unknown substance has a mass of 89.5 g. if 345.2 j of heat are required to heat the substance from 285 k to 305 k, what is the specific heat of the substance? use the formula q = mc,δt. j/g·k done
Answer
Explanation:
Step1: Identify given values
$q = 345.2\ J$, $m=89.5\ g$, $\Delta T=T_2 - T_1=305\ K - 285\ K = 20\ K$
Step2: Rearrange the formula for specific - heat
The formula is $q = mC\Delta T$. Rearranging for $C$ gives $C=\frac{q}{m\Delta T}$.
Step3: Substitute values into the formula
$C=\frac{345.2\ J}{89.5\ g\times20\ K}$ $C=\frac{345.2\ J}{1790\ g\cdot K}$ $C = 0.193\ J/g\cdot K$
Answer:
$0.193$