1.00 x 10^3 kg of clear liquid (specific heat
capacity = 5.11 x 10^2 J/kg-°C) at a temperature
of 15.0°C gains 3.33 x 10° J of heat. What is the
final temperature of the liquid? (Assume the
melting point is less than 15.0°C and the boiling
point is greater than 62.0°C.)

Where Q = heat gained by the liquid, c = specific heat capacity of the liquid, m = mass of the liquid, t₁ = Initial Temperature of the liquid, t₂ = Final temperature of the liquid.

Given: Q = 3.33 J, c = 511 J/kg°C, m = 1000 kg, t₁ = 15°C

Substitute these values into equation 1 and solve for t₂

3.33 = 511×1000(t₂-15)

3.33 = 511000(t₂-15)

(t₂-15) = 3.33/511000

(t₂-15) = 0.0000065

t₂ = 15+ 0.0000065

t₂ = 15.00000652 °C

1.00 x 10^3 kg of clear liquid (specific heat capacity = 5.11 x 10^2 J/kg-°C) at a temperature of 15.0°C gains 3.33 x 10° J of heat. What is the final temperature of the liquid? (Assume the melting point is less than 15.0°C and the boiling point is greater than 62.0°C.)

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1.00 x 10^3 kg of clear liquid (specific heat
capacity = 5.11 x 10^2 J/kg-°C) at a temperature
of 15.0°C gains 3.33 x 10° J of heat. What is the
final temperature of the liquid? (Assume the
melting point is less than 15.0°C and the boiling
point is greater than 62.0°C.)