A) \[2.0\operatorname{kc}\operatorname{a}\operatorname{l}/\operatorname{kg}\]
B) \[7\operatorname{kc}\operatorname{a}\operatorname{l}/\operatorname{kg}\]
C) \[3\operatorname{kc}\operatorname{a}\operatorname{l}/\operatorname{kg}\]
D) \[0.5\operatorname{kc}\operatorname{a}\operatorname{l}/\operatorname{kg}\]
Correct Answer: D
Solution :
As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from \[30{}^\circ C\] to \[25{}^\circ C\]) is same 2 minutes, therefore\[{{\left( \frac{dQ}{dt} \right)}_{water}}={{\left( \frac{dQ}{dt} \right)}_{liquid}}\] or,\[\frac{({{m}_{w}}{{C}_{w}}+W)\Delta T}{t}\frac{({{m}_{\ell }}{{C}_{\ell }}+W)\Delta T}{t}\] (W= water equivalent of the vessel) or, \[{{m}_{w}}{{C}_{w}}={{m}_{\ell }}{{C}_{\ell }}\] \[\therefore \]Specific heat of liquid, \[{{C}_{\ell }}=\frac{{{m}_{w}}{{C}_{w}}}{{{m}_{\ell }}}\] \[=\frac{50\times 1}{100}=0.5\,kcal/kg\]
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