In an experiment on the specific heat of a metal, a 0.20 kg block… - Vidyadip

Question

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at $150^{\circ} \mathrm{C}$ is dropped in a copper calorimeter (of water equivalent 0.025 kg ) containing $150 \mathrm{~cm}^3$ of water at $27^{\circ} \mathrm{C}$. The final temperature is $40^{\circ} \mathrm{C}$. 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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Answer

Mass of the metal, $m = 0.20kg = 200g$
Initial temperature of the metal, $T_1 = 150^\circ C$
Final temperature of the metal, $T_2 = 40^\circ C$
Calorimeter has water equivalent of mass, m’ = 0.025kg = 25g
Volume of water, $V = 150cm^3$
Mass (M) of water at temperature $T = 27^\circ C: 150 \times 1 = 150g$
Fall in the temperature of the metal:
$\triangle\text{T}=(\text{T}_1-\text{T}_2)=(150-40)=110^\circ\text{C}$
Specific heat of water, $\text{C}_\text{w}=\frac{4.186\text{J}}{\frac{\text{g}}{^\circ\text{K}}}$
Specific heat of the metal = C
Heat lost by the metal, $\theta=\text{m}\text{C}\triangle\text{T}\ ...(1)$
Rise in the temperature of the water and calorimeter system:
$\triangle\text{T}=(40-27)=13^\circ\text{C}$
Heat gained by the water and calorimeter system:
$\triangle\theta=\text{m}_1\text{C}_\text{w}\triangle\text{T}$
$=(\text{M}+\text{m}')\text{C}_\text{w}\triangle\text{T}'$
$(200\times\text{C}\times110)=(1(50+25))\times4.186\times13)$
$\frac{(175\times4.186\times13)}{(110\times200)}=0.43\text{jg}^{-1}\text{K}^{-1}$
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

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