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How much heat is required to warm 1.70 kg of sand from 28.0 ∘c to 100.0 ∘c?

User Akilesh
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2 Answers

4 votes

Final answer:

To warm 1.70 kg of sand from 28.0 °C to 100.0 °C, the amount of heat energy required is approximately 101,976 joules, calculated using the specific heat capacity of sand and the temperature change.

Step-by-step explanation:

The student is asking about the amount of heat transfer required to warm a certain mass of sand from one temperature to another. To calculate this, we use the formula:

Q = mcΔT

where:

Plugging in the values given:

Q = (1.70 kg)(830 J/kg°C)(100.0 °C - 28.0 °C)

Q = 1.70 kg × 830 J/kg°C × 72.0 °C

Q = 101,976 J

Therefore, the amount of heat energy required to warm 1.70 kg of sand from 28.0 °C to 100.0 °C is approximately 101,976 joules.

User Nunos
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8.5k points
5 votes
122 g * 4,186 (j/g*°c) * 23°c = 11745.916 j
User Mike
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8.1k points