An insulated container is used to hold 47.4 g of water at 31.1 °C. A sample of c

Question

An insulated container is used to hold 47.4 g of water at 31.1 °C. A sample of copper weighing 9.30 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.0°C. The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 0.385 J/(g·°C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Explanation / Answer

Ans. When Cu metal at 100.00C is put in water at 31.10C, Cu-metal loses some heat wand what gains same amount of heat in order to reach a thermal equilibrium temperature, T.

# . Amount of heat gained/lost by the samples is given by-

q = m s dT                            - equation 1

Where,

q = heat gained/lost

m = mass of sample

s1 = specific heat of sample

dT = Final temperature – Initial temperature

# Now,

Total heat lost by Cu-metal = Total heat gained by water

Or, -q (Cu-metal) = q (water)

Or, - [9.30 g x (0.385 J g-10C-1) x (T – 100.00C)] = 47.4 g x (4.184 J g-10C-1) x (T –31.10C)

Or, 3.5805 J 0C-1 x (100.00C - T) = 198.3216 J 0C-1 x (T –31.10C)

Or, (100.00C - T) = 198.3216 J 0C-1 x (T –31.10C) / (3.5805 J 0C-1)

Or, (T – 100.00C) = 55.38936 x (T –31.10C)

Or, 100.00C - T = 55.38936T – 1722.60910C

Or, 55.38936T + T = 100.00C + 1722.60910C

Or, 56.38936T = 1822.60910C

Or, T = 1822.60910C / 56.38936

Hence, T = 32.3220C           

Therefore, final temperature in the container = 32.320C

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