Question

a 2.74 g sample of a substance suspected of being pure gold is warmed to 72.1 °c and submerged into 15.2 g of water initially at 24.7 °c. the final temperature of the mixture is 26.3 °c. what is the heat capacity of the unknown substance? could the substance be pure gold?

Explanation:

Step1: Set up heat - transfer equation

$q_{substance}=q_{water}$, $q = mc\Delta T$.

Step2: Calculate $\Delta T$ values

$\Delta T_{substance}=26.3 - 72.1=-45.8^{\circ}C$, $\Delta T_{water}=26.3 - 24.7 = 1.6^{\circ}C$.

Step3: Calculate heat gained by water

$q_{water}=15.2\times4.18\times1.6\ J$.

Step4: Solve for $c_{substance}$

$c_{substance}=\frac{15.2\times4.18\times1.6}{2.74\times45.8}\ J/g^{\circ}C$.

Step5: Compare with gold's specific heat

Compare $c_{substance}$ with $c_{gold}=0.129\ J/g^{\circ}C$.

Answer:

1. First, use the principle of conservation of energy ($q_{lost}=q_{gained}$). The heat - transfer equation is $q = mc\Delta T$.

• The heat lost by the unknown substance is $q_{substance}=m_{substance}c_{substance}\Delta T_{substance}$.
• The heat gained by the water is $q_{water}=m_{water}c_{water}\Delta T_{water}$.
• We know that $m_{substance}=2.74\ g$, $T_{i - substance}=72.1^{\circ}C$, $T_f = 26.3^{\circ}C$, so $\Delta T_{substance}=T_f - T_{i - substance}=26.3 - 72.1=- 45.8^{\circ}C$.
• Also, $m_{water}=15.2\ g$, $c_{water}=4.18\ J/g^{\circ}C$, $T_{i - water}=24.7^{\circ}C$, so $\Delta T_{water}=T_f - T_{i - water}=26.3 - 24.7 = 1.6^{\circ}C$.

1. Since $q_{substance}=q_{water}$:

• $m_{substance}c_{substance}\Delta T_{substance}=m_{water}c_{water}\Delta T_{water}$.
• Substitute the known values: $2.74\ g\times c_{substance}\times(-45.8^{\circ}C)=15.2\ g\times4.18\ J/g^{\circ}C\times1.6^{\circ}C$.
• First, calculate the right - hand side: $15.2\ g\times4.18\ J/g^{\circ}C\times1.6^{\circ}C=15.2\times4.18\times1.6\ J\approx101.3\ J$.
• Then, solve for $c_{substance}$ from the left - hand side equation: $c_{substance}=\frac{15.2\times4.18\times1.6}{2.74\times45.8}\ J/g^{\circ}C$.
• $c_{substance}=\frac{101.2928}{125.492}\ J/g^{\circ}C\approx0.81\ J/g^{\circ}C$.
• The specific heat capacity of pure gold is $c_{gold}=0.129\ J/g^{\circ}C$. Since $0.81\ J/g^{\circ}C

eq0.129\ J/g^{\circ}C$, the substance is not pure gold.