Suppose 0.015 kg of steam (at 100.00 °C) is added to 0.50 kg of water (initially

Question

Suppose 0.015 kg of steam (at 100.00 °C) is added to 0.50 kg of water (initially at 11.0°C). The water is inside an aluminum cup of mass 60 g. The cup is inside a perfectly insulated calorimetry container that prevents heat flow with the outside environment. Find the final temperature of the water after equilibrium is reached. (Use the specific heats found in this table and the numerical data found in this table.)

____________°C

Specific heats table:-
http://www.webassign.net/bauerphys1/18-table-01.gif

Numerical data table:-
http://www.webassign.net/bauerphys1/18-table-02.gif

Thank you for your help

Explanation / Answer

According to principle of caloriemetry

heat given by steam = heat absorbed by water + container

ms * L + ms * c * (Ts  - T) = mw * c * (T - Tw) + mal * cal * (T - Tw)

ms * {L + c * (Ts  - T)} = (mw * c + mal * cal) * (T - Tw)

given mass of steam ms   = 0.015 kg

mass of water mw   = 0.50 kg

mass of cup mal   = 60 g = 0.06 kg

latent heat of steam L = 2260 * 103  J/kg

specific heat of water c = 4.19 * 103   J/kg- 0C

specific heat of aluminium cal   = 0.900 J/kg- 0C

Initial temperature of steam Ts  = 100 0C

Initial temperature of water Tw  = 11.0 0C

0.015 * {2260 * 103 + 4.19 * 103 * (100 - T)} = (0.50 * 4.19 * 103  + 0.06 * 0.900 * 103) * (T - 11.0)

2260 + 419 - 4.19 * T = 2.149 * T - 23.639

=> Equilibrium temperature T = (2679 + 23.639) / (2.149 + 4.19)

= 42.63 0C

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.