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How many grams of water are needed to absorb 456 J if its temperature goes from 22.7 to 98.3 Celsius?

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Eduard22slyidn

The mass of water needed to absorb 456 J is 1.44 g

We'll begin by calculating the change in the temperature of the water.

Initial temperature of water (T₁) = 22.7 °C

Final temperature (T₂) = 98.3 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 98.3 – 22.7

ΔT = 75.6 °C

  • Finally, we shall determine the mass of the water

Heat absorbed (Q) = 456 J

Change in temperature (ΔT) = 75.6 °C

Specific heat capacity of water (C) = 4.184 J/gºC

Mass of water (M) =?

Q = MCΔT

456 = M × 4.184 × 75.6

456 = M × 316.3104

Divide both side by 316.3104

M = 456 / 316.3104

M = 1.44 g

Therefore, the mass of the water is 1.44 g

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