Q) A) A 65.45-g piece of zinc metal at 75.0°C was dropped into a calorimeter con

Question

Q)

A) A 65.45-g piece of zinc metal at 75.0°C was dropped into a calorimeter containing 125 mL water at 19.9 °C. If the apparatus comes to thermal equilibrium at 22.2 °C, what is the heat capacity of the calorimeter?

In the same calorimeter, 5.00-g of NH4Cl was dissolved in 150.0 g of water at 20.7 °C. The temperature of the solution dropped to 18.9 °C as the ammonium chloride finished dissolving. B) What is the heat consumed by this reaction, q (J)?

C) What is the heat consumed by one mole dissolving, !H (kJ/mol)?

Explanation / Answer

So, the Heat gained by water is Q

Q = (mass)(specific heat)(temp change) = 125 X 4.18 X (22.2-19.9) = 1201.75J

Now Q = mass of metal X specific heat of etal X temperature hnage
Mass = 0.06545 kg
Specific heat of metal = 0.31
Q lost by metal = 0.06545 X S X (75-22.2) = 3.455 X 0.39 = 1.347 KJ = 1347 J

So heat absorbed by calrimeter = 1347-1201.75 = 145.25 J

so heat capacity of calorimeter = heat/temperature change = 145.25 / 2.3 = 63.15

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