Aluminum mass of metal sample is 90.5 g , mass of calorimeter cup is 7.9 g accep

Question

Aluminum mass of metal sample is 90.5 g , mass of calorimeter cup is 7.9 g accepted specific heat of the cup is 2 J/g degree C, mass of water inside calorimeter cup is 230.2,Txi=100 degree c, Twi=24.4 degree C and Txf=Twf= 30.8 degree c.

So find the computed specific heat, accepted specific heat and percent error.

FORMULA :

Cx= [Cw mw (Twf-Twi) + Ccal mcal (Twf-Twi) ] / [mx ( Txi - Txf) ]


please show all the work . what can we say about specific heat of metal comparison to specific heat of water. also metals are very good conductors of heat which is why the metal sample has to be rapaidly transferred to the calorimeter cup . what accout for this property ?

please show all the work . what can we say about specific heat of metal comparison to specific heat of water. also metals are very good conductors of heat which is why the metal sample has to be rapaidly transferred to the calorimeter cup . what accout for this property ?

Explanation / Answer

Mass of aluminum sample m1 = 90.5 g Accepted specific heat of aluminum Sa = 2 J/g degreeC Computed specific heat of aluminum Sc = ? Initial temperature of aluminum Txi = 100 degreeC Mass of copper calorimeter m3 = 7.9 g Specific heat of copper Sco = 0.386 J/g degreeC Mass of water m2 = 230.2 g Specific heat of water Sw = 4.186 J/g degreeC Initial temperature of water Twi = 24.4 degreeC Common temperature T = Txf = Twf = 30.8 degreeC From the law of calorimetry                          Heat lost by hot body (aluminum) = Heat gained by cold body (water)                                                         m1Sc(Txi - T) = [m2Sw+m3Sco] (T - Twi)                                                      98.4*Sc*69.2 = [(230.2*4.186) + (7.9*0.386)]*6.4                                                                         = 966.66 * 6.4                                                                       Sc = 0.9085J/g degreeC Percentage error = [|(0.9085 - 2)|/2]*100                            = 54.57 % Mass of aluminum sample m1 = 90.5 g Accepted specific heat of aluminum Sa = 2 J/g degreeC Computed specific heat of aluminum Sc = ? Initial temperature of aluminum Txi = 100 degreeC Mass of copper calorimeter m3 = 7.9 g Specific heat of copper Sco = 0.386 J/g degreeC Mass of water m2 = 230.2 g Specific heat of water Sw = 4.186 J/g degreeC Initial temperature of water Twi = 24.4 degreeC Common temperature T = Txf = Twf = 30.8 degreeC From the law of calorimetry                          Heat lost by hot body (aluminum) = Heat gained by cold body (water)                                                         m1Sc(Txi - T) = [m2Sw+m3Sco] (T - Twi)                                                      98.4*Sc*69.2 = [(230.2*4.186) + (7.9*0.386)]*6.4                                                                         = 966.66 * 6.4                                                                       Sc = 0.9085J/g degreeC Percentage error = [|(0.9085 - 2)|/2]*100                            = 54.57 % Mass of aluminum sample m1 = 90.5 g Accepted specific heat of aluminum Sa = 2 J/g degreeC Computed specific heat of aluminum Sc = ? Initial temperature of aluminum Txi = 100 degreeC Mass of copper calorimeter m3 = 7.9 g Specific heat of copper Sco = 0.386 J/g degreeC Mass of water m2 = 230.2 g Specific heat of water Sw = 4.186 J/g degreeC Initial temperature of water Twi = 24.4 degreeC Common temperature T = Txf = Twf = 30.8 degreeC From the law of calorimetry                          Heat lost by hot body (aluminum) = Heat gained by cold body (water)                                                         m1Sc(Txi - T) = [m2Sw+m3Sco] (T - Twi)                                                      98.4*Sc*69.2 = [(230.2*4.186) + (7.9*0.386)]*6.4                                                                         = 966.66 * 6.4                                                                       Sc = 0.9085J/g degreeC Percentage error = [|(0.9085 - 2)|/2]*100                            = 54.57 % (a) Specific heat of water/specific heat of metal = 4.607 So Specific heat of water is 4.067 times that of metal (b) That process is called Thermal conduction

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.