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