# What temperature will be established in the vessel if the mass of ice is 200 grams

What temperature will be established in the vessel if the mass of ice is 200 grams, the mass of water is 2 kg, and the initial water temperature is 30 degrees?

Q1 + Q2 = Q3.

We assume that ice is at a temperature of 0 ° C.

Q1 (ice melting) = λ * m1, where λ = 3.4 * 10 ^ 5 J / kg, m1 = 200 g = 0.2 kg.

Q2 (heating): Q2 = C * m1 * (t3 – t1), where C = 4200 J / (kg * K), t3 is the steady-state temperature, t1 = 0 ° C.

Q3 (cooling): Q2 = С * m2 * (t2 – t3), where m2 = 2 kg, t2 = 30 ° С.

λ * m1 + С * m1 * (t3 – t1) = С * m2 * (t2 – t3).

3.4 * 105 * 0.2 + 4200 * 0.2 * (t3 – 0) = 4200 * 2 * (30 – t3).

68000 + 840t3 = 252000 – 8400t3.

9240t3 = 184000.

t3 = 19.91 ° C.

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