Practice question
Question
A 0.2 kg aluminium block at 120° C is placed in 0.8 kg of water at 20° C in a 0.1 kg copper calorimeter at 20° C . What is the final temperature? (Specific heat of aluminium = 900 J kg^{-1 K^{-1, water = 4186 J kg^{-1 K^{-1, copper = 386 J kg^{-1 K^{-1 )
Explanation
Given:
A 0.2 kg aluminium block at 120° C is placed in 0.8 kg of water at 20° C in a 0.1 kg copper calorimeter at 20° C . What is the final temperature? (Specific heat of aluminium = 900 J kg^{-1 K^{-1, water = 4186 J kg^{-1 K^{-1, copper = 386 J kg^{-1 K^{-1 )
These values define the system as per NCERT data.
Formula:
Heat lost = Heat gained.
This is the standard NCERT relation for this phenomenon.
Substitution & Calculation:
0.2 × 900 × (120 - T) = (0.8 × 4186 + 0.1 × 386) × (T - 20) . 21600 - 180 T = (3348.8 + 38.6) × (T - 20) = 3387.4 T - 67748 . 21600 + 67748 = 3387.4 T + 180 T . 89348 = 3567.4 T Rightarrow T approx 25.04° C approx 25° C .
Result:
The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Discussion
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