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A 100 g chunk of aluminum is heated to 100 degrees C and placed in 500 g of water initially at...

Question:

A 100 g chunk of aluminum is heated to {eq}100^oC{/eq} and placed in 500 g of water initially at {eq}18.3^oC{/eq}. The final equilibrium temperature of the mixture is {eq}21.7^oC: \ (C_w= 1 \ cal/g-^oC=4186 \ J/kg-^oC){/eq}.

a. What is the specific heat of aluminum?
b. What is the total heat transferred to the water?

Heat Energy:

The energy needed to change the temperature of an object is known as that particular object's heat energy. It is directly proportional to the specific heat of the object and changes in the temperature. The expression of the heat energy is given as:

{eq}Q = mc \Delta T{/eq}, where:

  • m is the mass
  • c is the specific heat
  • {eq}\rm {\Delta T}{/eq} is the change in the temperature

Answer and Explanation:


We are given the following data:

  • Mass of the aluminum {eq}(m_{a}) = 100 \ g{/eq}
  • Mass of the water {eq}(m_{w}) = 500 \ g{/eq}
  • Initial temperature of the aluminium {eq}(T_{ia}) = 100^\circ C{/eq}
  • Initial temperature of the water {eq}(T_{iw}) = 18.3^\circ C{/eq}
  • Final temperature of the mixture {eq}(T) = 21.7^\circ C {/eq}


Question (a)


According to the principle of calorimetry, the heat lost is equal to the heat gain. Therefore:

{eq}\begin{align} \rm Q_{a} &= \rm Q_{w} \\ \rm m_{a}C_{a} (T_{ia} - T) &= \rm m_{w}C_{w} (T - T_{iw}) \\ \rm 100 \times C_{a} \times (100 -21.7) &= \rm 500 \times (4186 \ J/kg ^\circ C) (21.7 -18.3) \\ \rm C_{a} &= \rm \boxed{908.84 \ J/kg ^\circ C }\\ \end{align}{/eq}


Question (b)


The heat transferred to the water can be determined with the following relation:

{eq}Q_{w} = m_{w}c_{w} (T - T_{iw}) \\ Q_{w} =( 0.5 \ kg) \times (4186 \ J/kg ^\circ C) \times (2.7 -18.3) \\ \boxed{Q_{w} = 7116.2 \ J} {/eq}


Learn more about this topic:

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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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