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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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cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the surroundings is l
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Answer:

The system gains 126100 J

Explanation:

The heat can be calculated by the equation:

Q = nxCxΔT, where Q is the heat, C is the heat capacity,n is the number of moles and ΔT is the variation of temperature (final - initial). The number of moles is the mass divided by the molar mass, so:

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Q = 62.5x20.8x97

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