Question

a sample of tin (cp = 0.227 j/g·°c) is placed in a freezer. its temperature decreases from 15.0°c to -10.0°c as it releases 543 j of energy. what is the mass of the sample? round your answer to three significant figures. use the formula q = mc,δt.

Explanation:

Step1: Calculate the temperature change

The formula for temperature change is $\Delta T=T_{final}-T_{initial}$.
Given $T_{initial} = 15.0^{\circ}C$ and $T_{final}=- 10.0^{\circ}C$, so $\Delta T=-10.0 - 15.0=-25.0^{\circ}C$.

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

The heat - transfer formula is $q = mC_p\Delta T$. Rearranging for $m$ gives $m=\frac{q}{C_p\Delta T}$.
We know $q = 543\ J$ (heat released, so we can use its magnitude), $C_p=0.227\ J/g\cdot^{\circ}C$, and $\Delta T=- 25.0^{\circ}C$ (the magnitude of $\Delta T$ is $25.0^{\circ}C$ for calculation of mass).
Substitute the values into the formula: $m=\frac{543}{0.227\times25.0}$.
First, calculate $0.227\times25.0 = 5.675$.
Then, $m=\frac{543}{5.675}\approx95.7\ g$.

Answer:

$95.7$