For the specifc heat of metal equation: SH: -(M water)(SH water)(delta T water)

Question

For the specifc heat of metal equation: SH: -(M water)(SH water)(delta T water) + (M cal) (SH cal) (delta T cal) / (M metal) (delta T metal)
molar mass equation: MM= 25/ SH(j/G°C) DATA AND CALCULATIONS All calculation work is to be done on a separate sheet, neatly labeled with answers rounded to correct number of significant digits. Part A: Specific Heat Mass of calorimeter (cup and stirrer) Mass of calorimeter and water Mass of water Mass of metal Initial temperature of water and calorimeter Unknown number 403 1019 10ml lo 0ml(100 2380 25.8 Initial temperature of metal Final temperature of system 25.2 teusten SAT water qcal Specific Heat of metal O Approximate molar mass of metal MM =--25

Explanation / Answer

q (water ) = mass of water x sp. ht . of water x ( rise in tempeerature ie. delta t )

------------- = 100 x 1.00 x 1.4

------------- = 140 joules

q ( cal. ) = water equivalent* of calorimeter x rise in temperature

------------ = ( 1103.436 - 103.707 ) x 1.4

------------- = 1399.62 joules

Specific heat of metal = { q (water) + q (cal.) } / ( Mass of metal x rise in temperature )

-------------------------------- = ( 140 + 1399.62 ) / ( 30.348 x 1.4 )

--------------------------------- = 36.2372 joules / g o C

or, specific heat in J ( ie. kJ) = 0.03623 J / g o C

Therefore , approximate molar mass of the metal = 25 / 0.03623 --------------------------, as per given relation

---------------------------------------------------------------------- = 690.04

* Note that q (cal.) is calculated by using water equivalent of calorimeter using the first two readings in the experiment.   

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