Quantity of heat due to change in temperature:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{m}}{\mathbf{c}}{\mathbf{\u2206}}{\mathbf{T}}}$

Quantity of heat at freezing:

$\overline{){\mathbf{Q}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{w}}}{{\mathbf{L}}}_{{\mathbf{f}}}}$

Time, t:

$\overline{){\mathbf{t}}{\mathbf{=}}\frac{\mathbf{Q}}{\mathbf{P}}}$

The amount of heat released when the water temperature changes from 21°C to 0°C:

Q = (250)(4.186)[(273 + 21) - (273 + 0)] = 21976.5 J

Refer to the temperature versus time graph (Figure 2) when answering the questions in Parts C through F. A system consists of 250 g of water. The system, originally at TA = 21.0 °C, is placed in a freezer, where energy is removed from it in the form of heat at a constant rate. The figure shows how the temperature of the system takes t1 =10min=600s to drop to 0°C, after which the water freezes. Once the freezing is complete, the temperature of the resulting ice continues to drop, reaching temperature TB after an hour. The following specific heat and latent heat values for water may be helpful.

specific heat of ice (at 0∘C) | ci = 2.10 J/(g⋅K) |

latent heat of fusion (ice to water phase change at 0∘C) | Lf = 333.7 J/g |

specific heat of water (at 15∘C) | cw = 4.186 J/(g⋅K) |

latent heat of vaporization (water to steam phase change at 100∘C) | Lv = 2256 J/g |

specific heat of steam (at 110∘C) | cs = 2.01 J/(g⋅K) |

At what time t2 will the water be completely frozen so the temperature can begin to fall below 0^{∘}C?

Express your answer numerically in seconds.

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