• # question_answer A 1.0 g sample of substance A at $100{}^\circ C$ is added to 100 mL of ${{\text{H}}_{\text{2}}}\text{O}$at $25{}^\circ C$. Using separate 100 mL portions of${{\text{H}}_{\text{2}}}\text{O,}$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 $0.20\,J{{g}^{-1}}{{\,}^{o}}{{C}^{-1}}$ A) ${{T}_{C}}>{{T}_{B}}>{{T}_{A}}$            B) ${{T}_{B}}>{{T}_{A}}>{{T}_{C}}$C) ${{T}_{A}}>{{T}_{B}}>{{T}_{C}}$            D) ${{T}_{A}}={{T}_{B}}={{T}_{C}}$

Correct Answer: C

Solution :

Heat absorbed or evolved, $\Delta Q=ms\Delta t.$ Where,$m=$ mass of substance, s = specific heat$\Delta t=$ temperature difference It means $s\propto \frac{1}{\Delta t}.$ Higher the temperature of the given solution, lesser is the temperature difference. So higher is the specific heat. As, the specific heat order is A > B > C. So, the order of temperature of solution is A > B > C. Hence, the correct option is [c]. LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

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