To warm 2.0 L of tea (d = 1.01 g/mL; sp. heat =4.184 J/g o C) a cook places a 50

Question

To warm 2.0 L of tea (d = 1.01 g/mL; sp. heat =4.184 J/g oC) a cook places a 500 g block of stone (sp.heat = 2.449 J/g oC) at a temperature of 200oF into the teapot. Assuming that the tea was initiallyat 72 oF, what is the final temperature of the tea inoF?

a. 33 b. 91     c. 110 d. 140 e. 200 I already know the Answer Can someone PLease help explain! I knowit has something to do with Q= MCDeltaT To warm 2.0 L of tea (d = 1.01 g/mL; sp. heat =4.184 J/g degreeC) a cook places a 500 g block of stone (sp.heat = 2.449 J/g degreee C) at a temperature of 200 degree F into the teapot. Assuming that the tea was initiallyat 72 degreeF, what is the final temperature of the tea indegree F? a. 33 b. 91 c. 110 d. 140 e. 200 I already know the Answer Can someone PLease help explain! I knowit has something to do with Q= MCDeltaT

Explanation / Answer

   Volume of tea V = 2.0L                             = 2000 mL     density          d= 1.01 g/ mL mass of tea      m =V*d                              = 2000mL*1.01g/mL                              = 2020 g         sp.heat of tea s =4.184 J/g oC            initial temperature of tea T = 72oF      mass of block stone  m =500g       sp. heat of blockstone s = 2.449 J/g oC           initialtemperatue      T     = 200 oF    but we know loss of heat = gain of heat =Q                         q1 =q2           500g*2.449 J/g oC* (200oF -oF )   = 2020g*4.184 J/g oC*(oF - 72oF )                                        (200oF - oF ) / (oF -72oF ) =2020g*4.184 J/g oC/500g*2.449 J/g oC                                       bysolving this equation we get   the finaltemperature =    88.45oF     therefore answer B is correct.                             

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