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  • A metal block of mass  and initial temperature <img alt="T_1=300 \mathr

A metal block of mass m_1=0.5 \mathrm{~kg} and initial temperature T_1=300 \mathrm{~K} is placed in a container of water of mass m_2=2.0 \mathrm{~kg} and initial temperature T_2=280 \mathrm{~K}. After some time, both the metal block and water come to a common final temperature T_f=290 \mathrm{~K}. Given that the specific heat of the metal is c_{\text {metal }}=400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} and the specific heat of water is c_{\text {water }}=4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, calculate the specific heat capacity of the metal block.

Option: 1

41860 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}.


Option: 2

-41860 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}.


Option: 3

41860 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}.


Option: 4

21860 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}.


Answers (1)

best_answer

Given:

  • Mass of metal block m_1=0.5 \mathrm{~kg}
  • Initial temperature of metal block T_1=300 \mathrm{~K}
  • Mass of water m_2=2.0 \mathrm{~kg}
  • Initial temperature of water T_2=280 \mathrm{~K}
  • Final temperature T_f=290 \mathrm{~K}
  • Specific heat of metal c_{\text {metal }}=400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}
  • Specific heat of water c_{\text {water }}=4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}

Step 1: Calculate the heat gained by the metal block ( \left.Q_{\text {metal }}\right) :
            \begin{gathered} Q_{\text {metal }}=m_1 \cdot c_{\text {metal }} \cdot\left(T_f-T_1\right) \\ Q_{\text {metal }}=0.5 \mathrm{~kg} \cdot 400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot(290 \mathrm{~K}-300 \mathrm{~K}) \end{gathered}

Step 2: Calculate the heat lost by the water \left(Q_{\text {water }}\right) :
            \begin{gathered} Q_{\text {water }}=m_2 \cdot c_{\text {water }} \cdot\left(T_2-T_f\right) \\ Q_{\text {water }}=2.0 \mathrm{~kg} \cdot 4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot(280 \mathrm{~K}-290 \mathrm{~K}) \end{gathered}

Step 3: Since the heat gained by the metal is equal to the heat lost by the water at thermal equilibrium:

            Q_{\text {metal }}=Q_{\text {water }}

0.5 \mathrm{~kg} \cdot 400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot(290 \mathrm{~K}-300 \mathrm{~K})=2.0 \mathrm{~kg} \cdot 4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot(280 \mathrm{~K}-290 \mathrm{~K})
Step 4: Solve for the specific heat capacity of the metal ( \left.c_{\text {metal }}\right) :

            c_{\text {metal }}=\frac{2.0 \mathrm{~kg} \cdot 4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot(280 \mathrm{~K}-290 \mathrm{~K})}{0.5 \mathrm{~kg} \cdot(290 \mathrm{~K}-300 \mathrm{~K})}

Step 5: Calculate c_{\text {metal }} :
            c_{\text {metal }}=41860 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}
The specific heat capacity of the metal block is 41860 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}.

So, option A is correct

 

Posted by

chirag

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