Question

350 j of energy are released as ice cools from -5.0°c to -32°c. the specific heat of ice is 2.1 j/(g°c). what is the mass of the ice? specific heat equation: q = cp × m × δt

Explanation:

Step1: Calculate the temperature change

$\Delta T=T_{final}-T_{initial}=-32^{\circ}C - (-5^{\circ}C)=-27^{\circ}C$. The magnitude of $\Delta T$ is $27^{\circ}C$.

Step2: Rearrange the specific - heat formula to solve for mass

The specific - heat formula is $q = c_{p}\times m\times\Delta T$. Rearranging for $m$ gives $m=\frac{q}{c_{p}\times|\Delta T|}$.
We know that $q = 350J$, $c_{p}=2.1J/(g\times^{\circ}C)$, and $|\Delta T| = 27^{\circ}C$.
Substitute the values into the formula: $m=\frac{350J}{2.1J/(g\times^{\circ}C)\times27^{\circ}C}$.
First, calculate the denominator: $2.1J/(g\times^{\circ}C)\times27^{\circ}C = 56.7J/g$.
Then, $m=\frac{350J}{56.7J/g}\approx6.2g$.

Answer:

B. 6.2g