hhj10 10. Assume that 67 g 110 °C of steam is added to 689 g of water initially

Question

hhj10 10. Assume that 67 g 110 °C of steam is added to 689 g of water initially at 9°C. The water is inside an aluminum cup of mass 39 g. The cup is inside a perfectly insulating calorimetric container that prevents heat flow from the outside environment. Find the final temperature of the water after equilibrium is reached. 1cal=4.186J Specific heat of water: 4.19 kJ/(kg × K) or 1.00 cal/(g × K) Specific heat of ice: 2060 J/(kg × K) or 0.492 cal/(g × K) Specific heat of steam: 2010 J/kg × K) or 0.48 cal/(g K) Latent heat of fusion for water: 334 kJ/kg or 79.7 cal/g Latent heat of vaporization for water: 2260 kJ/kg or 539 cal/g Specific heat of aluminum: 922 J/(kg × K) or 0.22 cal/(g x K)

Explanation / Answer

Conservation of energy

Heat lost = Heat gained

m1 = 67 g = 0.067 g

m2 = 689 g = 0.689 kg

m3 = 39 g = 0.039 kg

m1 Cp1 ( 110 - 100) + m1 L1 + m1 Cp2 ( 100 - T) = m2 Cp2 ( T - 9) + m3 Cp3 (T - 9)

0.067 * 2010 * (110-100) + 0.067 * 2260*10^3 + 0.067 * 4190 *(100-T) = 0.689*4190*(T-9) + 0.039 * 922 * (T-9)

Solving above equation, we get

T = 64.66 o C

=========================================================

Final temperature = 64.66 o C

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