A piece of metal weighing 100 g is heated to 80° C and dropped into 1 kg of cold water in an insulated container at 15° C. If the final temperature of the water in the container is 15.69° C, the specific heat of the metal in J / g . ° C is

Question

A piece of metal weighing 100 g is heated to 80° C and dropped into 1 kg of cold water in an insulated container at 15° C. If the final temperature of the water in the container is 15.69° C, the specific heat of the metal in J / g . ° C is (A) 0.38 (B) 0.24 (C) 0.45 (D) 0.13

Tardigrade Admin · Accepted Answer

Given, weight of metal =100 g T2=15.69° C, T1=80° C Specific heat of water =4.184 J / g ° C Heat gained by 100 g of metal = Heat lost by 1000 g of water. We know, Q=m c Δ T ∴ 100 × x ×(80-15.69) =1000 × 4.184(15.69-15) =100 x(64.31)=2886.96 x =(2886.96/6431)=0.448 ≈ 0.45 J / g . ° C

Q. A piece of metal weighing $100 g$ is heated to $8 0^{\circ} C$ and dropped into $1 k g$ of cold water in an insulated container at $1 5^{\circ} C$ . If the final temperature of the water in the container is $15.6 9^{\circ} C$ , the specific heat of the metal in $J / g .^{\circ} C$ is

$0.38$

$0.24$

$0.45$

$0.13$

Q. A piece of metal weighing 100g is heated to 80∘C and dropped into 1kg of cold water in an insulated container at 15∘C. If the final temperature of the water in the container is 15.69∘C, the specific heat of the metal in J/g.∘C is

0.38

0.24

0.45

0.13

Given, weight of metal =100g T2​=15.69∘C,T1​=80∘C Specific heat of water =4.184J/g∘C Heat gained by 100g of metal = Heat lost by 1000g of water. We know, Q=mcΔT ∴100×x×(80−15.69) =1000×4.184(15.69−15) =100x(64.31)=2886.96 x=64312886.96​=0.448≈0.45J/g.∘C

Q. A piece of metal weighing $100 g$ is heated to $8 0^{\circ} C$ and dropped into $1 k g$ of cold water in an insulated container at $1 5^{\circ} C$ . If the final temperature of the water in the container is $15.6 9^{\circ} C$ , the specific heat of the metal in $J / g .^{\circ} C$ is

$0.38$

$0.24$

$0.45$

$0.13$