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How much heat is required to raise the temperature of 0.210 g of water from 19.2 ∘C to 32.0 ∘C?

Sagot :

Skandar

Data:

  • m = 0.210 g
  • T₀ = 19.2 °C + 273 = 292. 2 K
  • T = 32.0 °C  + 273 =  305 K
  • Ce = 4.18 J / GK

We apply the following formula

Q = mcₑΔT

    Q = mcₑ (T - T₀)

We substitute Our data in the formula and solve:

Q = 0.210 g * 4.18 J / g K (305 k - 292.2 k)

Q = 11.23 J

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