1.A 5.01 kg piece of granite with a specific heat of 0.803 J g -1 °C -1 and a te

Question

1.A 5.01 kg piece of granite with a specific heat of 0.803 J g-1 °C-1 and a temperature of 85.6 °C is placed into 2.00 L of water at 18.5 °C. When the granite and water come to the same temperature, what will the temperature be?

2. The combustion of methane (the chief component of natural gas) follows the equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H° for this reaction is -802.3 kJ. How many grams of methane must be burned to provide enough heat to raise the temperature of 197.3 mL of water from 17.23 °C to 40.34 °C?

Explanation / Answer

1) Let the common temperature attained be t°C. We shall employ the principle that heat lost by the granite block = heat gained by water.

Now, heat lost by the granite block = mG.CG.(85.6 – t)°C where mG denotes mass of the granite block in gm and CG is the specific heat of granite.

The water is at 18.5°C; density of water is 0.998 gm/cc at 18.5°C. Hence, mass of 2.00 L = 2000 cc is (2000 cc * 0.998 gm/cc) = 1996 gm.

The specific heat of water is 4.186 J/gm.°C.

Heat gained by water = mW.CW.(t – 18.5)°C.

According to the principle of thermochemistry,

mG.CG.(85.6 – t) = mW.CW.(t – 18.5)

or, (5.01 * 1000 gm)(0.803 J/gm.°C)(85.6 – t)°C = (1996 gm)(4.186 J/gm.°C)(t – 18.5)°C

or, (4023.03 J)(85.6 – t) = (8355.256 J)(t – 18.5)

or, 344,371.368 – 4023.03t = 8355.256t – 154,572.236

or, 344,371.368 + 154,572.236 = 8355.256t + 4023.03t

or, 498,943.604 = 12,378.286t

or, t = 40.308

The common temperature attained by both is 40.31°C (ans)

2) The density of water is again 0.998 gm/mL and the specific heat is 4.186 J/gm.°C. The mass of water is (197.3 mL).(0.998 gm/mL) = 196.905 gm

The heat gained by water is (196.905 gm)(4.186 J/gm.°C).(40.34 – 17.23)°C = 19048.286 J = 19.048 kJ.

Now the enthalpy of combustion of methane is given as -802.3 kJ. Enthalpy of combustion is always reported in kJ/mol, hence the correct value of the enthalpy here is -802.3 kJ/mol.

We ignore the negative sign (work only with numbers); also we assume that x mole of methane is burned to produce the heat that is responsible for raising the temperature of water. So,

(x mol)(802.3 kJ/mol) = 19.048 kJ

or, 802.3 x = 19.048

or, x = 0.0237

Molar mass of methane is 16 gm/mol. Hence, mass of 0.0237 mole of methane is

(0.0237 mol)(16 gm/mol) = 0.3792 gm

The amount of methane that must be combusted is 0.379 gm (ans)

Please note that I have not used the value of density of water as being 1.00 gm/cc or 1.00 gm/mL. The density of water is 1.00 gm/cc at 4°C (that is the maximum density of water) and the density is slightly lower at other temperatures. However, if you want, you can use the density of water as 1.00 gm/cc and the answers will be slightly different.

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