11.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

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Payal Gupta | Contributor-Level 10

5 months ago

11.14 Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, $T_{1}$ = 150 °C, Final temperature of the metal, $T_{2}$ = 40 °C

The water equivalent mass of the calorimeter, m’ = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass of water, M at T = 27 °C, = 150 $\times 1 = 150 g$

Fall in metal temperature, $?$ T = $T_{1} - T_{2}$ = 150 – 40 = 110 °C

Specific heat of water, $C_{w}$ = 4.186 J/g/ $°$ K

Let the specific heat of metal = C

Then, heat loss by the metal, $θ$ = mC $?$ T ……. (i)

Rise in the water of the calorimeter system $?$ T’ = 40 – 27

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11.14 Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, $T_{1}$ = 150 °C, Final temperature of the metal, $T_{2}$ = 40 °C

The water equivalent mass of the calorimeter, m’ = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass of water, M at T = 27 °C, = 150 $\times 1 = 150 g$

Fall in metal temperature, $?$ T = $T_{1} - T_{2}$ = 150 – 40 = 110 °C

Specific heat of water, $C_{w}$ = 4.186 J/g/ $°$ K

Let the specific heat of metal = C

Then, heat loss by the metal, $θ$ = mC $?$ T ……. (i)

Rise in the water of the calorimeter system $?$ T’ = 40 – 27 = 13°C

Heat gained by the water and calorimeter system; $θ'$ = (M + m’) $C_{w} ?$ T’ …… (ii)

Now heat lost by the metal = heat gained by the water and calorimeter system

mC $?$ T = (M + m’) $C_{w} ?$ T’

C = $\frac{(M + m ’) C_{w} ? T ’}{m ? T}$ = $\frac{(150 + 25) \times 4.186 \times 13}{200 \times 110}$ = 0.433 J/g/ $°$ K

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11.14 Mass of the metal, m = 0.20 kg = 200 gInitial temperature of the metal,  <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 150 °C, Final temperature of the metal,  <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 40 °CThe water equivalent mass of the calorimeter, m’ = 0.025 kg = 25 gVolume of water, V = 150 cm3Mass of water, M at T = 27 °C, = 150 <math><mo>×</mo><mn>1</mn><mo>=</mo><mn>150</mn><mi></mi><mi>g</mi></math>Fall in metal temperature,  <math><mo>? </mo></math> T = <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>-</mo><mi></mi><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 150 – 40 = 110 °CSpecific heat of water,  <math><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>w</mi></mrow></mrow></msub></math> = 4.186 J/g/ <math><mo>°</mo></math> KLet the specific heat of metal = CThen, heat loss by the metal,  <math><mi>θ</mi></math> = mC <math><mo>? </mo></math> T ……. (i)Rise in the water of the calorimeter system <math><mo>? </mo></math> T’ = 40 – 27 = 13°CHeat gained by the water and calorimeter system; <math><mi>θ</mi><mi>'</mi></math> = (M + m’) <math><mi></mi><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>w</mi></mrow></mrow></msub><mo>? </mo></math> T’ …… (ii)Now heat lost by the metal = heat gained by the water and calorimeter systemmC <math><mo>? </mo></math> T = (M + m’) <math><mi></mi><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>w</mi></mrow></mrow></msub><mo>? </mo></math> T’C = <math><mfrac><mrow><mrow><mo> (</mo><mi mathvariant="normal">M</mi><mi mathvariant="normal"></mi><mo>+</mo><mi mathvariant="normal"></mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">’</mi><mo>)</mo><mi></mi><msub><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mi>w</mi></mrow></mrow></msub><mo>? </mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">’</mi></mrow></mrow><mrow><mrow><mi>m</mi><mo>? </mo><mi mathvariant="normal">T</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mo> (</mo><mn>150</mn><mo>+</mo><mn>25</mn><mo>)</mo><mo>×</mo><mn>4.186</mn><mo>×</mo><mn>13</mn></mrow></mrow><mrow><mrow><mn>200</mn><mo>×</mo><mn>110</mn></mrow></mrow></mfrac></math> = 0.433 J/g/ <math><mo>°</mo></math> K

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