A 25.00 gram gold ingot and a 30.00 gram block of copper are placed in 100.00 gr

Question

A 25.00 gram gold ingot and a 30.00 gram block of copper are placed in 100.00 grams of water. If the initial temperatures of the gold, copper, and water were 95.0

A 25.00 gram gold ingot and a 30.00 gram block of copper are placed in 100.00 grams of water. If the initial temperatures of the gold, copper, and water were 95.0 A degree C, 85.0 A degree C, and 25.0 A degree C, respectively, what would the final temperature of the entire system be? The specific heats of gold, copper, and liquid water are 0.129, 0.387, and 4.18 J g-1 A degree C-1, respectively. 23.1 A degree C 28.2 A degree C 26.0 A degree C -27.1 A degree C 27.1 A degree C

Explanation / Answer

Gold mass = 25 g

copper mass = 30 g

water mass = 100 g

Ti gold = 95

Ti copper = 85

Ti water = 25

Final temperature of system

Cp gold = 0.129

Cp copper = 0.387

Cp water = 4.18

Let's first "cool" the gold

Qw = -Qg

mw*CPw*(Tf-Ti) = -mg*Cpg*(Tf-Ti)

100g*4.18*(Tf-25)=-25g*0.129*(Tf-95)

Solve for Tf

-129.6 (Tf-25) = (Tf-95)

-129.6Tf+3240 = Tf - 95

-128.6Tf = -3335

Tf = 25.9

Now let us use the copper!

Qw = - Qc

mw*Cpw*(Tf-Ti) = mc*Cpc*(Tf-Ti)

100g*4.18*(Tf-25.9) = -30g*0.387*(Tf-85)

-36 (tf-25.9) = tf-85

-36Tf+932.4 = Tf- 85

-37Tf = -1017.4

Tf = 27.49

Nearest answer shown: 27.1

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