Question

Use the References to access important values if needed for this question. An electric range burner weighing 6410 grams is turned off after reaching a temperature of 487.0°C, and is allowed to cool down to 22.6°C Calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 585.0 grams of water from 22.6°C to 81.3°C Answer J/gºC Submit Answer Retry Entire Group 3 more group attempts remaining

Answer 1

The complete step wise calculation of the specific heat capacity of the burner is given as below:-

It is given that Initial mass (mi) = 641.0g Final mass (ma) = 5850g Initi al temperature (T) = 487,0 °C Final Temperature (TQ) = 22.6°C last temperature te= 81.3°C Consider specific heat capacity =C, Then Heat given to the system= heat taken by the system. m, xc, x (Ti-Tf) = mqx Cq X (TE-T2) -

Putting all the values in the above equation , 641 XL1 X (487 -81-3) = 585xcqx (81.3 -22,6) - where, (= specific heat of water = 4.2 J/g °C Then from equation @ we have

641C1 405.7 = 58 59x(4.25 19°C) x (58.7) or c = 144925. 9 2600 53.7 J/2°c or a = 0.555 J/g °c So, the specific heat capacity of burner will be c= 0.56 J/g °C.