A beaker contains 200 gm of water- The heat capacity of the beaker is equal to that of 20 gm of water- The initial temperature of water in the beaker is 20-circ-C- If 440 gm of hot water 92-circ-C is poured in it- the final temperature- neglecting radiation loss- will be nearest to58-circ-C73-circ-C68-circ-C78-circ-C

Question

Toppr Admin · Accepted Answer

Let the final temperature of the system be $T$ .
Thus the heat absorbed by the cold water and beaker system= $200 \times 1 \times (T - 20) + 20 \times 1 \times (T - 20)$
Heat lost be the hot water= $440 \times 1 \times (92 - T)$
From conservation of heat energy,
$200 \times 1 \times (T - 20) + 20 \times 1 \times (T - 20)$ $= 440 \times 1 \times (92 - T)$
$⟹ T = 68^{\circ} C$

Let the final temperature of the system be T.Thus the heat absorbed by the cold water and beaker system=200×1×(T−20)+20×1×(T−20)Heat lost be the hot water=440×1×(92−T)From conservation of heat energy,200×1×(T−20)+20×1×(T−20) =440×1×(92−T)⟹T=68∘C