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  • A metal block of mass  = 0.2 kg and specific heat <img alt="c_{metal}" src="https://entrancecorner

A metal block of mass m_{1} = 0.2 kg and specific heat c_{metal}= 500 J/kg.K is placed in a container of water of mass m2 = 1.0 kg and specific heat c_{water} = 4186 J/kg·K. Both the metal block and water come to a common final temperature Tf = 320 K. Calculate the initial temperature T1 of the metal block.

Option: 1

320 K


Option: 2

340 K


Option: 3

390 K


Option: 4

230 K


Answers (1)

best_answer

Given:

Mass of metal block m_{1} = 0.2 kg

Specific heat of metal c_{metal} = 500 J/kg.K

Mass of water m_{2} = 1.0 kg

Specific heat of water c_{water} = 4186 J/kg.K

Final temperature Tf = 320 K

Step 1: Calculate the heat gained by the metal block (Qmetal)  to reach the final temperature  Tf:

        Q_{metal} = m_{1}. c_{metal}.(T_{f} - T_{1})

Step 2: Calculate the heat gained by the water (Qwater)  to reach the final temperature  Tf:

        Q_{water} = m_{2}. c_{water}.(T_{f} - T_{1})

Step 3: Since both the metal block and water comes to the same final temperature Tf, the heat gained by the metal is equal to the heat gained by the water:

        Q_{metal} = Q_{water}

   m_{1}. c_{metal}.(T_{f} - T_{1}) = m_{2}. c_{water}.(T_{f} - T_{1})

Step 4: Solve for the initial temperature T1 of the metal block:

m_{1}. c_{metal}.(T_{f} - T_{1}) = m_{2}. c_{water}.(T_{f} - T_{1})

0.2 kg * 500 J/kg.K * (320 K - T1) = 1.0 kg * 4186 J/kg.K * (320 K - T1)

Step 5: Solve for T1:

0.2 kg * 500 J/kg.K * (320 K - T1) = 1.0 kg * 4186 J/kg.K * (320 K - T1)

100 (320 - T1) = 4186 (320 - T1)

100 * 320 - 100 * T1 = 4186 * 320 - 4186 * T1

32000 - 100 * T1 = 1339520 - 4186 * T1

4086 * T1 = 1307520

T_{1} = \frac{1307520}{4086}

Step 6: Calculate T1:

                T1 \approx 320.33 K

The initial temperature T1 of the metal block is approximately 320.33 K. 

 

 

Posted by

Divya Prakash Singh

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