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The specific heat of liquid bromine is 0.226 J/g-K. How much heat (J) is required to raise the temperature of 10.0 mL of bromine from 25.00 °C to 27.30 °C? The density of liquid bromine: 3.12 g/mL

Respuesta :

Answer:

16.2 J

Explanation:

Step 1: Given data

  • Specific heat of liquid bromine (c): 0.226 J/g.K
  • Volume of bromine (V): 10.0 mL
  • Initial temperature: 25.00 °C
  • Final temperature: 27.30 °C
  • Density of bromine (ρ): 3.12 g/mL

Step 2: Calculate the mass of bromine

The density is equal to the mass divided by the volume.

ρ = m/V

m = ρ × V

m = 3.12 g/mL × 10.0 mL

m = 31.2 g

Step 3: Calculate the change in the temperature (ΔT)

ΔT = 27.30 °C - 25.00 °C = 2.30 °C

The change in the temperature on the Celsius scale is equal to the change in the temperature on the Kelvin scale. Then, 2.30 °C = 2.30 K.

Step 4: Calculate the heat required (Q) to raise the temperature of the liquid bromine

We will use the following expression.

Q = c × m × ΔT

Q = 0.226 J/g.K × 31.2 g × 2.30 K

Q = 16.2 J