when a block of metal of specific heat capacity 400j/kg.K and weighing 110 g is
ID: 1696967 • Letter: W
Question
when a block of metal of specific heat capacity 400j/kg.K and weighing 110 g is heated to 100 cantigrated and then quikly transfered to a calorimeter containing 200 g of a liquid at 10 cantigrated, the resulting temperature is 18 cantigrated. on repeating the experiment with 400 g of liquid in the same calorimeter and at the same initual temperature of 10 cantigrated, the resulting temperature is 14.5 cantigrated calculate from these observation the specific heat capacity of the liquid and the heat capacity of the calorimeterExplanation / Answer
Mass of the metal block m1 = 110 g = 0.11 Kg
Specific heat capacity of block S1 = 400 J/Kg K
Initial temperature of block t1 = 100 degrees C
Mass of the liquid m2 = 200 g = 0.2 Kg
Initial temperature of the liquid t2 = 10 degrees C
Resultent temperature t = 18 degrees C
Specific heat of the liquid S2 = ?
(a)
From the law of calorimetry
Heat lost by metal block = Heat gained by liquid + Heat gained by calorimeter
m1 S1 (t1 - t) = m2 S2 (t - t2) + mS(t - t2)
0.11 * 400 * 82 = 0.2 * S2 * 8 + m S * 8
0.2*S2 + m S = 451 -----------(1)
(b)
On repeating the same experiment with
m2 = 400 g = 0.4 Kg; t = 14.5 degrees C
and the remaining are same then
From the law of calorimetry
Heat lost by metal block = Heat gained by liquid + Heat gained by calorimeter
m1 S1 (t1 - t) = m2 S2 (t - t2) + mS(t - t2)
0.11* 400 * 85.5 = 0.4 * S2 * 4.5 + m S * 4.5
0.4*S2 + mS = 836 ---------------(2)
On subtracting (1) from (2) then we get
0.2 S2 = 385
Specific heat of liquid S2 = 1925 J Kg/K
Heat capacity of calorimeter (m = 1 Kg),
From (1),
0.2*S2 + m S = 451
385 + S = 451
Heat capacity of calorimeter S = 66 J/K
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.