Question #93a6f

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Question #93a6f

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Answer 1 · 2016-10-06T06:07:46

Given
$m_{w} \to Mass of water = its volume \times d e n s i t y$ $m_w->"Mass of water"="its volume"xx"density$
$= 25 c m^{3} \times 1 \frac{g}{c m^{3}} = 25 g$ $=25cm^3xx1g/(cm^3)=25g$

$t_{w} \to Initial temperature of water = 23^{\circ} C$ $t_w->"Initial temperature of water"=23^@C$

$m_{C d} \to Mass Cadmium = 65.5 g$ $m_(Cd)->"Mass Cadmium"=65.5g$

$t_{C d} \to Initial temperature of Cd = 100^{\circ} C$ $t_(Cd)->"Initial temperature of Cd"=100^@C$

$C_{w} \to Sp.heat capacity of water = 4.184 \frac{J}{g^{\circ} C}$ $C_w->"Sp.heat capacity of water"=4.184J/(g^@C)$

$C_{C d} \to Sp.heat capacity of Cd = 0.2311 \frac{J}{g^{\circ} C}$ $C_(Cd)->"Sp.heat capacity of Cd"=0.2311J/(g^@C)$

Let the final temperature of the system be $t^{\circ} C$ $t^@C$

So by calorimetric principle

$Heat lost by Cd = Heat gained by water$ $"Heat lost by Cd" = "Heat gained by water"$

$\Rightarrow m_{C d} \times C_{C d} \times (t_{C d} - t) = m_{w} \times C_{w} \times (t - t_{w})$ $=>m_(Cd)xxC_(Cd)xx(t_(Cd)-t)= m_wxxC_wxx(t-t_w)$

$\Rightarrow 65.5 \times 0.2311 \times (100 - t) = 25 \times 4.184 \times (t - 23)$ $=>65.5xx0.2311xx(100-t)= 25xx4.184xx(t-23)$

$\Rightarrow \frac{65.5 \times 0.2311}{25 \times 4.184} \times (100 - t) = (t - 23)$ $=>(65.5xx0.2311)/(25xx4.184)xx(100-t)= (t-23)$

$\Rightarrow 0.1447 \times (100 - t) = (t - 23)$ $=>0.1447xx(100-t)= (t-23)$

$\Rightarrow 14.47 - 0.1447 t = (t - 23)$ $=>14.47-0.1447t= (t-23)$

$\Rightarrow 1.1447 t = 37.47$ $=>1.1447t=37.47$

$\Rightarrow t = \frac{37.47}{1.1447} \approx {32.73}^{\circ} C$ $=>t=37.47/1.1447~~32.73^@C$

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Question #93a6f

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