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Answer to Question #91610 in General Chemistry for ivan

Question #91610
A student placed 45.0 g of water into a calorimeter and observed that the temperature was 20.0oC. At 60.0 g metal cylinder was heated to 92.0oC and then put into the calorimeter. The system was allowed to come to equilibrium. What was the final temperature in the calorimeter if the specfic heat of the metal cylinder was 0.150 cal/(g*oC) and the heat capacity of the calorimeter is 5.20 cal/oC?
Expert's answer

Write the heat balance equation:

@$c_mm_m(T_m - T_f) = c_wm_w(T_f - T_w) + c_c(T_f-T_c)@$

Open the brackets:

@$c_mm_mT_m - c_mm_mT_f = c_wm_wT_f - c_wm_wT_w + c_cT_f - c_cT_c@$

Separate known and unknown parts:

@$c_mm_mT_m + c_wm_wT_w + c_cT_c = c_wm_wT_f + c_cT_f + c_mm_mT_f@$

Take out the final temperature:

@$c_mm_mT_m + c_wm_wT_w + c_cT_c = T_f (c_wm_w + c_c + c_mm_m)@$

Find the final temperature

@$T_f = (c_mm_mT_m + c_wm_wT_w + c_cT_c) / (c_wm_w + c_c + c_mm_m)@$

Calculate the final temperature converting all temperature values into K

@$T_f = (0.15*60*365 + 1*45*293 + 5.2*293)/(1*45 + 5.2 + 0.15*60)@$

@$T_f = 303.9 K = 30.9^oC@$

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