Answer to Question #68713 in Molecular Physics | Thermodynamics for ujjwal
Let x represent the mass of the ice that will melt.
The heat supplied by the water cooling down is (4200 J/kgK) × (0.005 kg) × (30 K)
The heat received by the ice is (2050 J/kgK) × (0.005 kg) × (20 K) + (334 KJ/kg) × (x kg)
Assuming the system isolated, the heat lost by the warm water is equal to the heat received by the ice.
4200 x 0,005 x 30 = 2050 x 0.005 x 20 + 334.000x
Solving for x, obtaining x = 0.00127.
Therefore, 1.27 g of ice will melt, and the total mass of 8.73 g of water will remain at 0°C.
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