when a 39.0g metal block is placed in boiling...
when a 39.0g metal block is placed in boiling water, the metal absorbs 2445cal of energy. the blocks temperature changes from 30.0°c to 100°c. determine all unknowns about the metal block: m unit q unit tinitial unit tfinal unit δt unit cmetal unit
Answer
# Explanation:
## Step1: Identify given values
$m = 39.0g$, $q=2445cal$, $T_{initial}=30.0^{\circ}C$, $T_{final}=100^{\circ}C$
## Step2: Calculate $\Delta T$
$\Delta T=T_{final}-T_{initial}=100^{\circ}C - 30.0^{\circ}C=70.0^{\circ}C$
## Step3: Use heat - capacity formula
The formula for heat is $q = mc\Delta T$, where $q$ is heat energy, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is change in temperature. We can solve for $c_{metal}$:
$c_{metal}=\frac{q}{m\Delta T}=\frac{2445cal}{39.0g\times70.0^{\circ}C}$
$c_{metal}=\frac{2445cal}{2730g\cdot^{\circ}C}\approx0.8956cal/(g\cdot^{\circ}C)$
# Answer:
$m = 39.0g$, $q = 2445cal$, $T_{initial}=30.0^{\circ}C$, $T_{final}=100^{\circ}C$, $\Delta T = 70.0^{\circ}C$, $c_{metal}=0.8956cal/(g\cdot^{\circ}C)$