In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 150°C. Immediately, it is put into water of volume 150 cc at 27°C kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40°C. The specific heat of aluminium is: (take 4.2 Joule=1 calorie)

Question

In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 150°C. Immediately, it is put into water of volume 150 cc at 27°C kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40°C. The specific heat of aluminium is: (take 4.2 Joule=1 calorie) (A) 378 J/kg - °C (B) 315 J/kg - °C (C) 476 J/kg - °C (D) 434 J/kg - °C

Tardigrade Admin · Accepted Answer

Q given = Q used ⇒ m 1 C 1 Δ T 1=( m 2+ w ) C 2 Δ T 2 ⇒ 0 . 2 0 × C 1 ×( 1 5 0 - 4 0 )=( 1 5 0 × 1 0 -3) C 1= 4 3 4 ~ y k g -1 k -1

$378 J / k g -^{\circ} C$

$315 J / k g -^{\circ} C$

$476 J / k g -^{\circ} C$

$434 J / k g -^{\circ} C$

Q given = Q used
$\Rightarrow m_{1} C_{1} Δ T_{1} = (m_{2} + w) C_{2} Δ T_{2}$
$\Rightarrow 0.20 \times C_{1} \times (150 - 40) = (150 \times 1 0^{- 3})$
$C_{1} = 434 y k g^{- 1} k^{- 1}$

378J/kg−∘C

315J/kg−∘C

476J/kg−∘C

434J/kg−∘C

Q given = Q used ⇒m1​C1​ΔT1​=(m2​+w)C2​ΔT2​ ⇒0.20×C1​×(150−40)=(150×10−3) C1​=434 ykg−1k−1

$378 J / k g -^{\circ} C$

$315 J / k g -^{\circ} C$

$476 J / k g -^{\circ} C$

$434 J / k g -^{\circ} C$

Q given = Q used
$\Rightarrow m_{1} C_{1} Δ T_{1} = (m_{2} + w) C_{2} Δ T_{2}$
$\Rightarrow 0.20 \times C_{1} \times (150 - 40) = (150 \times 1 0^{- 3})$
$C_{1} = 434 y k g^{- 1} k^{- 1}$