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In an experiment, a piece of metal is heated to a high temperature and then placed in an insulated container with water. The initial temperature of the water is $T_{i}=25^{\circ} \mathrm{C}$ and the initial temperature of the metal is $T_{m}=150^{\circ} \mathrm{C}$ . After thermal equilibrium is reached, the final temperature of the water and metal mixture is $T_{f}=30^{\circ} \mathrm{C}$ . The mass of the metal is $m=0.2 \mathrm{~kg}$ , and the mass of water is $M=0.5 \mathrm{~kg}$ . The specific heat of water is $c_{w}=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$ .

The specific heat cm of the metal is:
Option: 1
$100 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$

Option: 2
$250.5 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$

Option: 3
$437.5 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$

Option: 4
$500 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$

Answers (1)

best_answer

The heat lost by the hot metal is equal to the heat gained by the cold water. This can be expressed using the formula:

$m \cdot c_{m} \cdot\left(T_{m}-T_{f}\right)=M \cdot c_{w} \cdot\left(T_{f}-T_{i}\right)$

Given:

$\begin{aligned} m & =0.2 \mathrm{~kg} \\ T_{m} & =150^{\circ} \mathrm{C} \\ M & =0.5 \mathrm{~kg} \\ T_{i} & =25^{\circ} \mathrm{C} \\ T_{f} & =30^{\circ} \mathrm{C} \\ c_{w} & =4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C} \end{aligned}$

Solve for cm :

$c_{m}=\frac{M \cdot c_{w} \cdot\left(T_{f}-T_{i}\right)}{m \cdot\left(T_{m}-T_{f}\right)}$

Substitute the given values:

$c_{m}=\frac{0.5 \mathrm{~kg} \cdot 4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C} \cdot\left(30^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)}{0.2 \mathrm{~kg} \cdot\left(150^{\circ} \mathrm{C}-30^{\circ} \mathrm{C}\right)}$

Calculate the value of cm :

$c_{m}=\frac{21000 \mathrm{~J} / \mathrm{kg}}{2400 \mathrm{~K}}=437.5 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$

So, the specific heat cm of the metal is approximately $437.5 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$ .

Posted by

shivangi.bhatnagar

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