a sample of tin ($c_p = 0.227 j/gcdot^{circ}c$) is placed in a freezer. its temperature decreases from…

a sample of tin ($c_p = 0.227 j/gcdot^{circ}c$) is placed in a freezer. its temperature decreases from $15.0^{circ}c$ to $- 10.0^{circ}c$ as it releases 543 j of energy. what is the mass of the sample? round your answer to three significant figures.\nuse the formula $q = mc_pdelta t$.

a sample of tin ($c_p = 0.227 j/gcdot^{circ}c$) is placed in a freezer. its temperature decreases from $15.0^{circ}c$ to $- 10.0^{circ}c$ as it releases 543 j of energy. what is the mass of the sample? round your answer to three significant figures.\nuse the formula $q = mc_pdelta t$.

Answer

Explanation:

Step1: Identify values

$q = - 543\ J$ (negative as heat is released), $C_p=0.227\ J/g\cdot^{\circ}C$, $\Delta T=T_2 - T_1=-10.0^{\circ}C - 15.0^{\circ}C=-25.0^{\circ}C$

Step2: Rearrange the formula

We have $q = mC_p\Delta T$, so $m=\frac{q}{C_p\Delta T}$

Step3: Substitute values

$m=\frac{-543\ J}{0.227\ J/g\cdot^{\circ}C\times(- 25.0^{\circ}C)}$ $m=\frac{-543}{0.227\times(-25.0)}\ g$ $m=\frac{-543}{-5.675}\ g$ $m = 95.7\ g$

Answer:

$95.7$