Question #56148

in a different experiment it is observed that 252J of heat must be absorbed to raise the temp of 50.0g of Ni(s) from 20.0 C to 31.4 C

calculate the specific heat of Ni in both J/g Degree C and cal/g Degree C

calculate the specific heat of Ni in both J/g Degree C and cal/g Degree C

Expert's answer

The specific heat of Ni is defined:

C = Q/(m∆T), where Q – the absorbed heat, m – the mass and ∆T – the change of temperature.

C = 252 J/(50.0 g × [31.4 oC – 20.0 oC]) = 0.442 J/(g· oC)

1 calorie equal 4.18 J, therefore Q = 252 J = 252 J/4.18 = 60.29 cal.

Hence, C = 60.29 cal/(50.0 g × [31.4 oC – 20.0 oC]) = 0.106 cal/(g· oC)

C = Q/(m∆T), where Q – the absorbed heat, m – the mass and ∆T – the change of temperature.

C = 252 J/(50.0 g × [31.4 oC – 20.0 oC]) = 0.442 J/(g· oC)

1 calorie equal 4.18 J, therefore Q = 252 J = 252 J/4.18 = 60.29 cal.

Hence, C = 60.29 cal/(50.0 g × [31.4 oC – 20.0 oC]) = 0.106 cal/(g· oC)

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