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musickatia [10]
3 years ago
10

If a student performs an exothermic reaction in a calorimeter, how does the calculated value of ΔH (Hcalc) differ from the actua

l value (Hactual) if the heat exchanged with the calorimeter is not taken into account?
Chemistry
1 answer:
devlian [24]3 years ago
7 0

Answer:

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

Explanation:

The amount of heat changed during this process at a fixed pressure is termed Enthalpy

enthalpy change ∆H = ∆E + P∆V

∆E = internal energy change

P = fixed pressure

∆V = change in volume

When energy is absorbed during reaction, it is called endothermic reaction.

Endothermic reaction carried out in the calorimeter and enthalpy change for the reaction. Since we have that

q(surrounding) = q(solution)+q(calorimeter)

Therefore, q(calorimeter) > 0(endothermic).

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

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A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a l
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Answer:

T_{C} = -4.2°C

T_{H} = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

|\frac{Q_{C} }{Q_{H} }| = \frac{T_{C} }{T_{H} }

Similarly,

|Q_{H}| = |Q_{C}| + |W_{S}|

During winter, the value of |T_{H}| = 20°C = 273.15 + 20 = 293.15 K and |W_{S}| = 1.5 kW. Therefore,

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|\frac{W_{S} }{Q_{H} }| = 1 - \frac{T_{C} }{T_{H} }

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Further simplification,

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During summer, T_{C} = 25°C = 273.15+25 = 298.15 K, and |W_{S}| = 1.5 kW. Therefore,

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Similarly,

|\frac{W_{S} }{Q_{C} }| = \frac{T_{H} }{T_{C} } - 1

1.5/0.75*(T_{H} - 298.15) = (T_{H}/298.15

Further simplification,

T_{H} = 49.4°C

4 0
4 years ago
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