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

Be very clear with your answer please

Thank you for your help

Explanation / Answer

The mass of the steam ms = 0.015kg

the mass of water mw = 0.5kg

the mass of aluminium cup mAl = 0.06kg

from calorimetry the heat lost by steam is equal to heat gained by water and cup

      ms Lv + ms c( 100 - T) = mw c(T - 11) + mAl cAl ( T- 11)

    (0.015)(2260*10^3 J/kg) + (0.015)(4186)(T - 11) = (0.5)(4186)(T - 11) + (0.06)(900)(T - 11)

     33900 + 62.79T - 690.69 = 2093T - 23023 + 54T - 594

      56826.31 = 2084.21 T

therefore the final temperature

           T = 27.265 deg C

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