\(500 \,g\) of water and \(100\, g\) of ice at \(0^{\circ}C\) are in a calorimeter whose water equivalent is \(40\, g\). \(10\, g\) of steam at \(100^\circ C\) is added to it. Then water in the calorimeter is : (Latent heat of ice \(= 80\, cal/g\), Latent heat of steam \(=540\, cal/g\))