Question

A silver cube with an edge length of 2.33 cm and a gold cube with an edge length of 2.71 cm are both heated to 81.9 ∘C and placed in 106.5 mL of water at 19.9 ∘C . What is the final temperature of the water when thermal equilibrium is reached?Matter Specific Heat Density (gr/ml3)Au 0.1256 19.3Ag 0.238 10.5H2O 4.184 1.00

Answer 1

Answer:

The final temperature of the water is 32 °C

Explanation:

Step 1: Given data

The silver cube has an edge length of 2.33 cm

The gold cube has an edge length of 2.71 cm

Both are heated to 81.9 °C and then placed in 106.5 mL water at 19.9 °C

Silver has a specific heat of 0.238 J/g*K and a density of 19.3 g/mL

Gold has a specific heat of 0.1256 J/g*K  and a density of 10.49 g/mL

Water has a specific heat of 4.184 J/g*K and a density of 1 g/mL

Step 2: Calculate the volume of silver, gold

V(Ag) = 2.33³ = 12.6493 cm³

V(Au) = 2.71 ³ = 19.9025 cm³

Step 3: Calculate the mass of silver, gold and water

m(Ag) = 12.6493 cm³ * 10.49 g/cm³ = 132.69 grams

m(Au) = 19.9025 cm³ * 19.3 g/cm³ = 384.118 grams

m(w) = 106.5 mL * 1 g/mL = 106.5 grams

Step 4: Calculate the final temperature of water

q(Ag) + q(Au) = - q(w)

m(Ag)*ΔT * C(Ag) + m(Au) *ΔT * C(Au) = - m(w) * ΔT * C(w)

m(Ag)*(T2-T1,Ag)*C(Ag) + m(Au)*(T2-T1,Au)*C(Au) = - m(w)*(T2-T1,w)*C(w)

m(Ag)*T2*C(Ag) - m(Ag)*T1,Ag *C(Ag) + m(Au)*T2*C(Au) - m(Au)*T1,Au*C(Au) = - m(w) * T2* C(w) + m(w)*T1(w)*C(w)

m(Ag)*T2*C(Ag) + m(Au)*T2*C(Au) + m(w)*T2*C(w) = m(Ag)*T1,Ag *C(Ag) + m(Au)*T1,Au*C(Au) + m(w) * T1,w* C(w)

T2(m(Ag)*C(Ag) + M(Au) * C(Au) + m(w) * C(w)) = m(Ag)*T1,Ag * C(Ag) + m(Au) *T1,Au  * C(Au) + m(w)*T1,w * C(w)

T2 = [ m(Ag)*T1,Ag*C(Ag) + m(Au) * T1,Au*C(Au) + m(H20) * T1,(w)*C(w) ]/ [m(Ag)*C(Ag) + m(Au)*C(Au)+ m(w)*C(w)]

T2= [132.69 * (81.9) * (0.1256) + 384.11 * (81.9) * (0.238) + 106.5 * (19.9) * (4.184)] / [132.69*(0.1256) +384.11 *( 0.238) + 106.5*(4.184)]

T2 = 32 °C

The final temperature of the water is 32 °C

Answer 2

At the equilibrium point, the heat lost by the silver and gold cubes is equal

to the eat gained by the water.

The final temperature of the water is approximately 29.33°C.

Reasons:

The given parameters are;

Edge length of silver cube = 2.33 cm

Edge length of gold cube = 2.71 cm

Volume of water = 106.5 mL = 106.5 cm³

Water temperature = 19.9°C

$\begin{tabular}{|l|c|c|}\underline{Substance}&\underline{Specific \ heat \ (J/(g \cdot ^{\circ}C)}&\underline{Density \ (g/cm^3)}\\Gold&0.1256&19.3\\Silver&0.2386&10.5\\Water&4.184&1.00\end{array}\right]$

Required:

The final temperature of the water

Solution:

Mass of the silver = 2.33³ × 10.5 ≈ 132.82

Mass of the silver, m$_s$ ≈ 132.82 g

Mass of the gold = 2.71³ × 19.3 ≈ 384.12

Mass of the gold, $m_g$ ≈ 384.12 g

Mass of the water = 106.5 cm³ × 1.00 g/cm³ = 106.5 g

Let T₂ represent the final equilibrium temperature. The heat released by

the metals absorbed by the water, therefore;

m$_s$ × $c_s$ × (T$_s$ - T₂) + $m_g$ × $c_g$ × (T$_g$ - T₂)

Where;

$c_s$ = The specific heat capacity of silver

$c_g$ = The specific heat capacity of gold

0.2386 × 132.82×(81.9 - T₂) + 0.1256× 384.12×(81.9 - T₂) = 4.184×106.5×(T₂ - 19.9)

6546.7849356 - 79.936324·T = 445.596·T - 8867.3604

$T_2 = \dfrac{15414.1453356}{525.532324} \approx 29.33$

The final temperature of the water, T₂ ≈ 29.33°C

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