A beaker contains 200 g of water. The heat capacity of beaker is equal to that of 20 g of water. The initial temperature of water in the beaker is 20° C. If 440 g of hot water at 92° C is poured in, the final temperature, neglecting radiation loss, will be

Question

A beaker contains 200 g of water. The heat capacity of beaker is equal to that of 20 g of water. The initial temperature of water in the beaker is 20° C. If 440 g of hot water at 92° C is poured in, the final temperature, neglecting radiation loss, will be (A) 58° C (B) 68° C (C) 73° C (D) 78° C

Tardigrade Admin · Accepted Answer

Heat lost =(440 g )(1. cal . g -1 ° C -1)(92-T)° C Heat gained =(200+20) g(1 operatornamecalg-1 ° C -1) ⋅(T-20) ° C ⇒ 440(92-T)=220(T-20) or 2(92-T)=T-20 or (2 × 92)-2 T=T-20 ⇒ 3 T=(184+20) T=(204/3) ⇒ T=68° C

Q. A beaker contains $200 g$ of water. The heat capacity of beaker is equal to that of $20 g$ of water. The initial temperature of water in the beaker is $2 0^{\circ} C$ . If $440 g$ of hot water at $9 2^{\circ} C$ is poured in, the final temperature, neglecting radiation loss, will be

$5 8^{\circ} C$

$6 8^{\circ} C$

$7 3^{\circ} C$

$7 8^{\circ} C$

Q. A beaker contains 200g of water. The heat capacity of beaker is equal to that of 20g of water. The initial temperature of water in the beaker is 20∘C. If 440g of hot water at 92∘C is poured in, the final temperature, neglecting radiation loss, will be

58∘C

68∘C

73∘C

78∘C

Heat lost =(440g)(1 cal g−1∘C−1)(92−T)∘C Heat gained =(200+20)g(1calg−1∘C−1)⋅(T−20)∘C ⇒440(92−T)=220(T−20) or 2(92−T)=T−20 or (2×92)−2T=T−20 ⇒3T=(184+20) T=3204​ ⇒T=68∘C

Q. A beaker contains $200 g$ of water. The heat capacity of beaker is equal to that of $20 g$ of water. The initial temperature of water in the beaker is $2 0^{\circ} C$ . If $440 g$ of hot water at $9 2^{\circ} C$ is poured in, the final temperature, neglecting radiation loss, will be

$5 8^{\circ} C$

$6 8^{\circ} C$

$7 3^{\circ} C$

$7 8^{\circ} C$