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Evgen [1.6K]
3 years ago
14

80 grams of water releases 2550 joules of heat. What is the temperature change of the water? Heat capacity of water is 4.2 J/g°C

.
Chemistry
1 answer:
Nitella [24]3 years ago
3 0

Answer:

7.59°C

Explanation:

Using the formula;

Q = m × c × ∆T

Where;

Q = amount of heat (joules)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

According to this question;

m = 80g

c = 4.2 J/g°C.

Q = 2550J

∆T = ?

Using Q = m × c × ∆T

∆T = Q/mc

∆T = 2550 ÷ (80 × 4.2)

∆T = 2550 ÷ 336

∆T = 7.589°C

The change in temperature of water = 7.59°C.

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A steel calorimeter has a volume of 75.0 mL and is charged with Oxygen gas to a
EastWind [94]

Answer:

7.78×10¯³ mole

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 75 mL

Pressure (P) = 255 kPa

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Next, we shall convert 75 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

75 mL = 75 mL × 1 L / 1000 mL

75 mL = 0.075 L

Next, we shall convert 22.5 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Temperature (T) = 22.5 °C

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Temperature (T) = 295.5 K

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Volume (V) = 0.075 L

Pressure (P) = 255 kPa

Temperature (T) = 295.5 K

Gas constant (R) = 8.314 KPa.L/Kmol

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Divide both side by 2456.787

n = 19.125 / 2456.787

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