120 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are

Question

120 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 1100 grams and initial temperature 22° C (the heat capacity of aluminum is 0.9 J/gram/K).

(a) After a short time, what is the temperature of the water?
Tfinal = C

(b) What simplifying assumptions did you have to make?

-Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.
-The heat capacities for both water and aluminum hardly change with temperature in this temperature range.
-The thermal energy of the aluminum doesn't change.
-The thermal energy of the water doesn't change.


(c) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 26432 J of work, and the temperature of the water and pan increases to 77.8° C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?
Q = J

Explanation / Answer

mass of the water m_w = 120 g initial temperature of water T_w = 100 ° C heat capacity of water c_w =4.2 J/g/K final temp =T loss of heat by water Q_w = m_wc_w(T_w1-T) ......... (1) mass of aluminum pan m_Al = 1100 g intial temp T_Al = 22 ° C heat capacity c_Al = 0.9 J/g/K final temp =T gain of heat Q_Al = m_Alc_Al(T-T_Al1) ....... (2) -------------------------------------------------------------------------------------------- a) if no heat transmits to surroundings in the short time then loss of heat by water is equal to heat gain by pan. i.e.,Q_w =Q_Al from (1) and (2), (120 gm)(4.2 J/g/K)(100 ° C -T)=(1100 gm)(0.9 J/g/K)(T-22 ° C) by solving we get T=48.35 ° C -------------------------------------------------------------------------------------------------- b) we cosider -Energy transfer between the system (water plus pan) and the surroundings was negligible during this time -------------------------------------------------------------------------------------------------- c) on heating, the temperature of the water and pan increases to 77.8° C heat gain by water is Q_w3 = m_wc_w(77.8 ° C-48.35 ° C) Q_w3=(120 gm)(4.2 J/g/K )(29.45 °C) =14862.80 J heat gain by aluminum pan Q_Al4 =m_Alc_Al(77.8-48.35° C) Q_Al4=(1100 gm)(0.9 J/g/K)(29.45°C)=29155.5 J total heat gain Q=Q_w3+Q_Al4 =44018.3 J the energy due to stirrer Q_st = 26432 J the energy supplied from stove Q' = Q-Q_st Q'=44018.3 J-26432 J=17586.3 J Q' =17586.3 J

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