a cube of aluminum ($c_{p}=0.897 j/gcdot k$) has a temperature of 325 k. it is placed in a beaker of cold…

a cube of aluminum ($c_{p}=0.897 j/gcdot k$) has a temperature of 325 k. it is placed in a beaker of cold water. if the aluminum cools to 301 k as the water absorbs 815.7 j of heat, what is the mass of the cube? use the formula $q = mc_{p}delta t$.
Answer
Explanation:
Step1: Identify given values
$q = 815.7\ J$, $C_p=0.897\ J/g\cdot K$, $T_1 = 325\ K$, $T_2=301\ K$, $\Delta T=T_2 - T_1$.
Step2: Calculate $\Delta T$
$\Delta T=301\ K - 325\ K=- 24\ K$
Step3: Rearrange the heat - transfer formula for mass
From $q = mC_p\Delta T$, we can solve for $m$: $m=\frac{q}{C_p\Delta T}$.
Step4: Substitute values and calculate mass
$m=\frac{815.7\ J}{0.897\ J/g\cdot K\times(- 24\ K)}=\frac{815.7}{- 21.528}\ g\approx - 37.9\ g$. Mass cannot be negative in this context, we take the absolute value. So $m = 37.9\ g$.
Answer:
$37.9$