This problem is easily solved by the methods of calorimetry, that is, conservation of energy. It is necessary to remember that the ice must meit completely before its temperature (and that of the water and the cup) will rise above 0C. The energy balance equation, Qhot : _Qcoid' for this system is given by the followmg. Q melt + Q warm melted Q warm initial + Q warm cup 2 _Q cool lead the ice ice to final temp water to final temp to nal temp to final temp The heat of fusion of water is if = 3.33 x 105 J/kg and the specic heat of water is c = 4,186 J/kg - \"C. The specic heat of lead is CPb = 128 J/kg - \"C and water that of copper is cc\" = 387 Jy'kg - ElC. When we rewrite the energy balance equation in terms of the egwlibrium temperature T1,, the speCIfic heats, heats of fusion, mass of ice m mass of melted ice m mass of the original water m mass of the copper cup m and the mass of the piece of ice' melted ice' original water' cup' lead mph, where TLPb is the initial temperature of the lead, we have water(Tf 0C) + m original C ICE Water = 'mpbcpbfrf' Tum)- T, 00c) + m Tf 00c) water( cu pCCu( miceLf + m melted c Substituting the given data gives (0.029 kg)(3.33 x 105 J/kg) + (0.029 kg)(4,186 J/kg - 0c)(12c 0C) 4 (0.195 kg)(4,186 J/kg - ac)(12c 00c) + (0.100 kg)(387 J/kg - c)(12c 0C) = empbuze J/kg - Dc)(12c 7 moc) or, entering term by term in the same order as above, 0.8058 3: Your response differs from the correct answer by more than 10%. Double check your calculations. K 104] -x Your response differs from the correct answer by more than 10%. Double check your calculations. K 103 J -x Your response differs from the correct answer by more than 10%. Double check your calculations. K 104] + 4.644 y x 1021 = mpb ( 10?52 x Your response differs from the correct answer by more than 10%. Double check your calculations. K 104 J/kg)