Question

A piece of metal of 850 K is dropped into 1 kg of water at 300 K. If equilibrium temperature of water is 350 K then the heat capacity of the metal expressed in \(\frac{J}{K}\) is \((\text{Sp. heat of water} = 1 \frac{\text{cal}}{\text{g}^{\circ}\text{C}})\)

• A. \(420\)
• B. \(240\)
• C. \(100\)
• D. \(24\)

Hint:

Heat lost by the metal equals heat gained by the water.

Answer 1

The Correct Option is A

Step 1: Approach
Write the energy balance with heat capacities.

Step 2: Balance
Water has heat capacity $1000\times4.2=4200$ J/K. Water rises by 50 K, the metal falls by 500 K, a factor of 10 larger.

Step 3: Equation
$C_{metal}\times500=4200\times50$, so $C_{metal}=\dfrac{4200}{10}=420$ J/K.

Step 4: Answer
Option (A).

Final Answer:
The water gains 210000 J while the metal falls 500 K, so its heat capacity is 420 J per K, option (A). \[ \boxed{420\ \text{J/K}} \]