A) \[40{}^\circ C\]
B) \[60{}^\circ C\]
C) \[0{}^\circ C\]
D) \[50{}^\circ C\]
Correct Answer: C
Solution :
1 kg of ice at 0°C is converted into water at 0°C. Heat taken by ice. =mL \[=1\times 336=336k\]joule Heat given to water to raise its temp from 0°C to \[T{}^\circ C.\text{ }T{}^\circ C\]is temperature of mixture. \[336+mc\Delta T={{m}_{w}}.c\Delta T\] \[336kJ+1\times 4.2kJ\] \[(T-O)=1\times 4.2\times kJ(80-T)\] \[336\times {{10}^{3}}+4.2\times {{10}^{3}}(T)=4.2\times {{10}^{3}}(80-T)\] \[336\times 4.2T=336-4.2T\] \[T={{0}^{o}}C\]
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