A piece of metal at 850 K is dropped into 1 kg water at 300 K. If the equilibrium temperature of water is 350 K then the heat capacity of the metal, expressed in J K⁻¹ is (1 cal = 4.2 J) (a) 420 (b) 240 (c) 100 (d) No Solution

Answer: (d). Heat gained by water = 1 × 4200 × 50 = 2.1 × 10⁵ J Metal falls by 850 - 350 = 500 K: C × 500 = 2.1 × 10⁵ C = 420 J K⁻¹, which is option (a). The given key marks (d) No Solution.

A piece of metal at 850 K is dropped into 1 kg water at 300 K. If the… | MHT CET 2023 (10 May, Shift 1)

Q20MHT CET 2023 · 10 MAY · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A piece of metal at $850\text{ K}$ is dropped into $1\text{ kg}$ water at $300\text{ K}$ . If the equilibrium temperature of water is $350\text{ K}$ then the heat capacity of the metal, expressed in $\text{J K}^{-1}$ is $(1\text{ cal} = 4.2\text{ J})$

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Answer: (d)

Heat gained by water $= 1 \times 4200 \times 50 = 2.1 \times 10^5\text{ J}$
Metal falls by $850 - 350 = 500\text{ K}$ : $C \times 500 = 2.1 \times 10^5$
$C = 420\text{ J K}^{-1}$ , which is option (a).
The given key marks (d) No Solution.

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