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 $q = mc_pdelta t$. \boxed{} g done

Explanation:

Step1: Determine ΔT

ΔT = T_final - T_initial = -10.0°C - 15.0°C = -25.0°C (the negative sign indicates energy release).

Step2: Rearrange the formula for mass (m)

From \( q = mC_p\Delta T \), we get \( m=\frac{q}{C_p\Delta T} \).
We know \( q = - 543\space J \) (negative because energy is released), \( C_p = 0.227\space J/g\cdot^\circ C \), and \( \Delta T=- 25.0^\circ C \).

Step3: Substitute the values into the formula

\( m=\frac{-543\space J}{0.227\space J/g\cdot^\circ C\times(- 25.0^\circ C)} \)
First, calculate the denominator: \( 0.227\times(-25.0)= - 5.675 \)
Then, \( m=\frac{-543}{-5.675}\approx95.7\space g \) (rounded to three significant figures)

Answer:

\( 95.7 \)