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A glass of water with a mass of 0.32 kg at 20 C is to be cooled to 0 C by dropping ice cubes at 0...

Question:

A glass of water with a mass of 0.32 kg at 20{eq}^{\circ} {/eq}C is to be cooled to 0{eq}^{\circ} {/eq}C by dropping ice cubes at 0{eq}^{\circ} {/eq}C into it. The latent heat of fusion of ice is 334 kJ/kg, and the specific heat of water is 4.18 {eq}kJ/kg\cdot^{\circ}C {/eq}. The amount of ice that needs to be added is

(a) 32 g

(b) 40 g

(c) 80 g

(d) 93 g

(e) 110 g

Law of Conservation of Energy:

The law of conservation of energy states that the energy cannot be created or destroyed but it can be changed from one form to the other form. For example, the heat energy can be converted into mechanical energy and vice versa.

Answer and Explanation:

{eq}\textbf {Given:} {/eq}

  • Initial temperature of water, {eq}T_{1w}=20^\circ \text {C} {/eq}
  • Final temperature of water, {eq}T_{2w}=0^\circ \text {C} {/eq}
  • Mass of water, {eq}m_w=0.32\ \text {kg} {/eq}

Latent heat of fusion of ice, {eq}h_{fg}=334\ \text {kJ/kg} {/eq}

Specific heat of water, {eq}C_{pw}=4.18\ \text {kJ/kg.}^\circ \text {C} {/eq}


Write the energy balance equation.

{eq}Q_{lost}=Q_{gained}\\ m_ih_{fg}=-m_wC_{pw}(T_{2w}-T_{1w})\\ m_i\times 334=-0.32\times 4.18\times (0-20)\\ m_i=0.080096\ \text {kg}\\ =80.1\ \text {g} {/eq}

Thus, the correct option is (c). That, is the amount of ice required is {eq}\bf \color {navy} {80\ \text {g}} {/eq}.


Learn more about this topic:

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What is Energy Conservation? - Definition, Process & Examples

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6
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