A piece of ice with a mass of m2 = 250 g at a temperature of t2 = 0 ° C is thrown into water
A piece of ice with a mass of m2 = 250 g at a temperature of t2 = 0 ° C is thrown into water, occupying a volume of V = 4 liters and heated to a temperature of t1 = 20 0C. Determine the temperature of the water after the ice has melted.
Data: V (volume of water taken) = 4 l = 0.004 m3; t1 (initial water temperature) = 20 ºС; m2 (mass of a piece of ice) = 250 g = 0.25 kg; t2 – initial temperature of a piece of ice (t2 = 0 ºС).
Constants: according to the condition Sv (specific heat capacity of water) = 4200 J / (kg * ºС); λ (specific heat of melting of ice) = 3.35 * 10 ^ 5 J / kg; ρw (water density) = 1000 kg / m3.
1) Heat expended to melt a piece of ice: Q = λ * m2 = 3.35 * 10 ^ 5 * 0.25 = 83.75 * 10 ^ 3 J.
2) Change in temperature of taken water: Q = Sv * V * ρw * (t1 – t), whence we express: t = t1 – Q / (Sv * V * ρv) = t1 – Q / (Sv * V * ρw) = 20 – (83.75 * 10 ^ 3) / (4200 * 0.004 * 1000) = 15 ºС.
Answer: After the ice melts, the water will have a temperature of 15 ºС; after equilibrium is established – 14.1 ºС.