# A 3.8 kg gold bar at 96 C is dropped into 0.30 kg of water at 24 C. What is the final temperature?Assume the specific heat of gold is 129 J/kg*C.

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### 2 Answers

A 3.8 kg gold bar at 96 C is dropped into 0.30 kg of water at 24 C. The temperature of water is less than that of the gold bar. Heat passes from the bar of gold to the water till the temperature of both of them is the same. The specific heat of gold is 129 J/kg*C and the specific heat of water is 4186 J/kg*C.

Let the final equilibrium temperature be T. The heat lost by gold to reach this temperature is 3.8*129*(96 - T). The heat gained by the water to reach the final temperature is 0.3*4186*(T - 24). The two have to have to be equal.

=> 3.8*129*(96 - T) = 0.3*4186*(T - 24)

=> 47059.2 - 490.2*T = 1255.8*T - 30139.2

=> T(490.2 + 1255.8) = 47059.2 + 30139.2

=> T = 77198.4/1746

=> T = 44.2 C

**The final temperature is 44.2 C**

Mass of gold bar = 3.8 Kg

Mass of water = 0.3 Kg

Specific Heat of gold = 129vJ/Kg C

Specific Heat of Water = 4184 J/Kg C

Initial Temperature of gold bar = 96 C

Initial Temperature of water = 24 C

Final Equilibrium Temperatur = x C

Rise in water Temperature = x-24 C

Fall in gold temperature = 96-x C

Heat gain/Loss = mass*specific heat* temperature change

Heat gain by water = 0.3*4184*(x-24) = 30124.8-1255.2*x J

Heat loss by gold = 3.8*129*(96-x) = 490.2*x-47059.2 J

Heat Gain = Heat loss

490.2*x-47059.2 = 30124.8-1255.2*x J

1745.4*x = 77184

x = 44.2

**Final Temperature = 44.2 C**