Tnt Drink

should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value. IDENTIFY: Apply Q = mcΔT to each object. The net heat flow Qsystem for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the “‘water” by w. T is the final equilibrium temperature. cw = 4190 J/kg ⋅ K. ΔTK = ΔTC . EXECUTE: (a) Qsystem = 0 gives mmCm ΔTm…

should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value. IDENTIFY: Apply Q = mcΔT to each object. The net heat flow Qsystem for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the “‘water” by w. T is the final equilibrium temperature. cw = 4190 J/kg ⋅ K. ΔTK = ΔTC . EXECUTE: (a) Qsystem = 0 gives mmCm ΔTm…

should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value. IDENTIFY: Apply Q = mcΔT to each object. The net heat flow Qsystem for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the “‘water” by w. T is the final equilibrium temperature. cw = 4190 J/kg ⋅ K. ΔTK = ΔTC . EXECUTE: (a) Qsystem = 0 gives mmCm ΔTm…

should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value. IDENTIFY: Apply Q = mcΔT to each object. The net heat flow Qsystem for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the “‘water” by w. T is the final equilibrium temperature. cw = 4190 J/kg ⋅ K. ΔTK = ΔTC . EXECUTE: (a) Qsystem = 0 gives mmCm ΔTm…

should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value. IDENTIFY: Apply Q = mcΔT to each object. The net heat flow Qsystem for the system (man, soft drink) is zero. SET UP: The mass of 1.00 L of water is 1.00 kg. Let the man be designated by the subscript m and the “‘water” by w. T is the final equilibrium temperature. cw = 4190 J/kg ⋅ K. ΔTK = ΔTC . EXECUTE: (a) Qsystem = 0 gives…

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