A) 2.0 kcal/kg
B) 7 kcal/kg
C) 3 kcal/kg
D) 0.5 kcal/kg
Correct Answer: D
Solution :
[d] \[\frac{\left( {{m}_{w}}{{C}_{m}}+W \right)\Delta T}{t}=\frac{\left( {{m}_{\ell }}{{C}_{\ell }}+W \right)\Delta T}{t}\] (W=water equivalent of the vessel) or \[{{m}_{w}}{{C}_{w}}={{m}_{\ell }}{{C}_{\ell }}\] \[\therefore \] Specific heat of liquid, \[C=\frac{{{m}_{w}}{{C}_{w}}}{{{m}_{\ell }}}\] \[=\frac{50\times 1}{100}=0.5kcal/kg\]
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