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Answer :

The required temperature is `546^@C`. Solution :

Let the volume at `0^@C` = V ml. <br> `V_1 = VmL, V_2 = 3 mL` <br>`T_1 = 0^@` or `273 K, T_2 = ?` <br> by charles law, <br> ` V_1/T_1 = V_2/T_2` or `V/273 =(3V)/T_2` <br> or `T_2 = (3xx273) = 819 K` <br> or `T_2 = (819-273)^@C = 546^@C`