# A silver block, initially at 58.5°C, is submerged into 100.0 of water at 24.8°C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.2 °C. What is the mass of the silver block?

Jan 28, 2018

$= 77.6 g$

#### Explanation:

1. Given the water's temperature, its density at that particular temperature can be found through interpolation; i.e., ( Density of water at various temperature are given in the attached table ). http://www.engineeringcivil.com/relative-density-of-water.html
${\rho}_{w a t e r} \text{@"24.8^oC=0.9971244" g/ml}$
2. Through the density formula, mass of the water in the container can be computed as shown below:
$\text{mass"(m)=rhoxx"volume(V)}$
$m = \frac{0.9971244 g}{\cancel{m l}} \times 100.0 \cancel{m l}$
$m = 99.71 g$
3. Now, find the heat absorbed during the submersion of the silver. Knowing that the thermal equilibrium is at ${26.2}^{o} C$, the ${Q}_{\text{gained}}$ by the water is computed as follows: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html
${Q}_{\text{gained}} = {m}_{w} C {p}_{w} \Delta T$
$w h e r e$:
$\Delta T = {T}_{f} - T i$=change in temperature(final and initial)
${Q}_{\text{gained}} = m C {p}_{w} \left({T}_{f} - {T}_{i}\right)$
${Q}_{\text{gained}} = \left(99.71 g\right) \left(\frac{4.186 J}{g {\cdot}^{o} C}\right) \left(26.2 - 24.8\right) {\cdot}^{o} C$
${Q}_{\text{gained}} = 583.753 J \approx 584 J$
4. Take note that ${Q}_{\text{lost")=Q_("gained}}$; so that, this relationship can be used to find the mass of the silver being cooled in the water. The Cp of the silver is given in the attached table. Rearrange the formula as needed. http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html
${Q}_{l o s t} = m C p \Delta T$
Note: ${Q}_{\text{lost}} = - Q$
$m = \frac{- Q}{C p \Delta T}$; where: $\Delta T = {T}_{f} - {T}_{i}$
$m = \frac{- Q}{\left(C p\right) \left({T}_{f} - {T}_{i}\right)}$
$m = \frac{- 584 \cancel{J}}{\left(\frac{0.233 \cancel{J}}{g \cdot \cancel{^ o C}}\right) \left(26.2 - 58.5\right) \cdot \cancel{^ o C}}$
$m = \frac{- 584 g}{- 7.53}$
$m = 77.6 g$