A 1.0 g sample of substance A at 100 o C is added to 100 mL of H 2 O at 25 o C . Using separate 100 mL portions of H 2 O , the procedure is repeated with substance B and then with substance C. How will the final temperatures of the water compare? Substance Specific Heat A 0.60 J g - 1 o C - 1 B 0.40 J g - 1 o C - 1 C 0.20 J g - 1 o C - 1

Question

A 1.0 g sample of substance A at 100 o C is added to 100 mL of H 2 O at 25 o C . Using separate 100 mL portions of H 2 O , the procedure is repeated with substance B and then with substance C. How will the final temperatures of the water compare? Substance Specific Heat A 0.60 J g - 1 o C - 1 B 0.40 J g - 1 o C - 1 C 0.20 J g - 1 o C - 1, T C > T B > T A, T B > T A > T C, T A > T B > T C, T A = T B = T C

MARKS App · Accepted Answer

Heat absorbed or evolved, Δ Q = m s Δ t Where, m = mass of substance, s = specific heat Δ t = temperature difference It me Solution: s ∝ 1 Δ t Higher the temperature of given solution, lesser is the temperature difference. So higher is the specific heat. As, specific heat order is A > B > C. So, order of temperature of solution is A > B > C.

Substance	Specific Heat
A	$0.60 J {g^{- 1}}^{o} C^{- 1}$
B	$0.40 J {g^{- 1}}^{o} C^{- 1}$
C	$0.20 J {g^{- 1}}^{o} C^{- 1}$

Search any question & find its solution

Looking for more such questions to practice?