Question

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A 10.5-kg block of ice has a temperature of -22.1 °c. The pressure is one atmosphere. The block absorbs 6.99 x of heat. What is the final temperature in degrees Celsius of the liquid 10 the tolerance is +/-2% Click if you would like to show work for this question: open Show works

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Answer 1

(1) The final temperature in degree celsius of a liquid water will be given as :

we know that, $\Delta$Q = m_ice C_ice (T_f - T₀)

(6.99 x 10⁶ J) = (10.5 kg) (2093 J/kg ⁰C) [T_f + (22.1 ⁰C)]

(6.99 x 10⁶ J) = (21976.5 J/⁰C) [T_f + (22.1 ⁰C)]

T_f + (22.1 ⁰C) = [(6.99 x 10⁶ J) / (21976.5 J/⁰C)]

T_f = [(318.06 ⁰C) - (22.1 ⁰C)]

T_f = 295.96 ⁰C

T_f$\approx$ 296 ⁰C

(2) The mass of water that vaporizes which will be given as :

(Heat lost by mercury) = (Heat gained by water) + (Heat needed for phase change)

$\Delta$Q_lost = $\Delta$Q_gained + $\Delta$Q_vaporization

m_Hg C_Hg$\Delta$T = m_w C_w$\Delta$T + m $\Delta$H

(3.96 kg) (140 J/kg ⁰C) [(204 - 100) ⁰C] = (0.212 kg) (4186 J/kg ⁰C) [(100 - 93.1) ⁰C] + m (333 x 10³ kg)

(57657.6 J) - (6123.2 J)] = m (333 x 10³ J/kg)

m = (51534.4 J) / (333 x 10³ J/kg)

m = 0.154 kg