How much heat is required to warm 1.50 kg of sand from 30.0 ∘c to 100.0∘c? the heat capacity of sand is 0.84 j⋅g−1⋅k−1?

8.8 × 10⁴ J

Explanation:

Step 1: Given data

mass of sand (m): 1.50 kg × (10³ g/ 1 kg) = 1.50 × 10³ g

Initial temperature: 30.0°C + 273.15 = 303.2 K

Final temperature: 100.0°C + 273.15 = 373.2 K

Change in temperature (ΔT): 373.2 K - 303.2 K = 70.0 K

Heat capacity of sand (c): 0.84 J/g.K

Step 2: Calculate the heat (Q)

We can calculate the required heat using the following expression.

Q = c × m × ΔT

Q = (0.84 J/g.K) × (1.50 × 10³ g) × 70.0 K = 8.8 × 10⁴ J

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.