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Question Number 10286 by Tawakalitu ayo mi last updated on 02/Feb/17

2000g of water at 100°C is poured into a copper calorimeter 150g of water at 10°C. The temperature of the mixture is 45°C. Calculate the thermal capaity of tbe vessel. specific heat capacity of copper = 400 specific heat capacity of water = 4200

$$\mathrm{2000g}\:\mathrm{of}\:\mathrm{water}\:\mathrm{at}\:\mathrm{100}°\mathrm{C}\:\mathrm{is}\:\mathrm{poured}\:\mathrm{into}\:\mathrm{a}\: \\ $$$$\mathrm{copper}\:\mathrm{calorimeter}\:\mathrm{150g}\:\mathrm{of}\:\mathrm{water}\:\mathrm{at}\:\mathrm{10}°\mathrm{C}. \\ $$$$\mathrm{The}\:\mathrm{temperature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mixture}\:\mathrm{is}\:\mathrm{45}°\mathrm{C}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{thermal}\:\mathrm{capaity}\:\mathrm{of}\:\mathrm{tbe}\:\mathrm{vessel}. \\ $$$$\mathrm{specific}\:\mathrm{heat}\:\mathrm{capacity}\:\mathrm{of}\:\mathrm{copper}\:=\:\mathrm{400} \\ $$$$\mathrm{specific}\:\mathrm{heat}\:\mathrm{capacity}\:\mathrm{of}\:\mathrm{water}\:=\:\mathrm{4200} \\ $$

Answered by ridwan balatif last updated on 02/Feb/17

To answer this question, you must using Azaz black m_1 =2kg, △T_1 =100−45=55^o C, c=4200J/kg^o C(it′s releasing heat)→water m_2 =0.15kg,△T_2 =45−10=35^o C, c=4200J/kg^o C(it′s accept heat)→water in calorimeter Q_(releasing) =Q_(accept) m_1 ×c×△T_1 =C_(calorimeter) ×△T_2 +m_2 ×c×△T_2 2×4200×55=C×35+0.15×4200×35 C=12570J/^o C

$$\mathrm{To}\:\mathrm{answer}\:\mathrm{this}\:\mathrm{question},\:\mathrm{you}\:\mathrm{must}\:\mathrm{using}\:\mathrm{Azaz}\:\mathrm{black} \\ $$$$\mathrm{m}_{\mathrm{1}} =\mathrm{2kg},\:\bigtriangleup\mathrm{T}_{\mathrm{1}} =\mathrm{100}−\mathrm{45}=\mathrm{55}^{\mathrm{o}} \mathrm{C},\:\mathrm{c}=\mathrm{4200J}/\mathrm{kg}^{\mathrm{o}} \mathrm{C}\left(\mathrm{it}'\mathrm{s}\:\mathrm{releasing}\:\mathrm{heat}\right)\rightarrow\mathrm{water} \\ $$$$\mathrm{m}_{\mathrm{2}} =\mathrm{0}.\mathrm{15kg},\bigtriangleup\mathrm{T}_{\mathrm{2}} =\mathrm{45}−\mathrm{10}=\mathrm{35}^{\mathrm{o}} \mathrm{C},\:\mathrm{c}=\mathrm{4200J}/\mathrm{kg}^{\mathrm{o}} \mathrm{C}\left(\mathrm{it}'\mathrm{s}\:\mathrm{accept}\:\mathrm{heat}\right)\rightarrow\mathrm{water}\:\mathrm{in}\:\mathrm{calorimeter} \\ $$$$\mathrm{Q}_{\mathrm{releasing}} =\mathrm{Q}_{\mathrm{accept}} \\ $$$$\mathrm{m}_{\mathrm{1}} ×\mathrm{c}×\bigtriangleup\mathrm{T}_{\mathrm{1}} =\mathrm{C}_{\mathrm{calorimeter}} ×\bigtriangleup\mathrm{T}_{\mathrm{2}} +\mathrm{m}_{\mathrm{2}} ×\mathrm{c}×\bigtriangleup\mathrm{T}_{\mathrm{2}} \\ $$$$\mathrm{2}×\mathrm{4200}×\mathrm{55}=\mathrm{C}×\mathrm{35}+\mathrm{0}.\mathrm{15}×\mathrm{4200}×\mathrm{35} \\ $$$$\mathrm{C}=\mathrm{12570J}/^{\mathrm{o}} \mathrm{C} \\ $$

Commented by Tawakalitu ayo mi last updated on 02/Feb/17

I really appreciate your effort sir. God bless you.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\: \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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