Test Index

Thermodynamics Part 5

Section: Physics

Question : 23 of 86

Marks: +1, -0

-10^{\circ}\,\text{C}

25^{\circ}\,\text{C}

\_\_\_\_

^{\circ}\,\text{C}

=2100\,\text{J}/\text{Kg}\,{}^{\circ}\text{C}

=4200\,\text{J}/\text{kg}\,{}^\circ\text{C}

=3.36 \times 10^{5}\,\text{J}/\text{Kg}

Solution:

Concept:

Calorimetry: The heat lost by the warm water must equal the total heat gained by the ice to warm up, melt, and then warm up again as water.

Explanation:

Let the final equilibrium temperature be

T_f

^\circ\text{C}

The heat lost by the water cooling from

25^\circ\text{C}

T_f

is:

Q_{\text{loss}} = m_w c_w (25 - T_f)

100 \times 4200 \times (25 - T_f)

10500000 - 420000 T_f

The heat gained by the ice occurs in three stages:

1. Warming ice from

-10^\circ\text{C}

0^\circ\text{C}

Q_1 = m_i c_i (0 - (-10))

10 \times 2100 \times 10

210000

2. Melting the ice at

0^\circ\text{C}

to water at

0^\circ\text{C}

Q_2 = m_i L_f

= $10 \times

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