A mass of 50 g of a certain metal at 150-C is immersed in 100 g of water at 11-C- The final temperature is 20 -C- Calculate the specific heat capacity of the metal- Assume that the specific heat capacity of water is 4-2 J g-1 K-1-

Heat lost by the metal = Heat gained by water m_m S_m Δ T_m = m_w S_w Δ T_w 50 × s_m × (150 20) = 100 × 4.2 × (20 11) s_m = 0.582 Jg^ 1K^ 1 ...

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Standard IX

Physics

Calorimetry

A mass of 50 ...

Question

A mass of $50 g$ of a certain metal at $150^{\circ} C$ is immersed in $100 g$ of water at $11^{\circ} C$ . The final temperature is $20^{\circ} C$ . Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$ .

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Solution

Heat lost by the metal = Heat gained by water
$⟹ m_{m} S_{m} Δ T_{m} = m_{w} S_{w} Δ T_{w}$
$50 \times s_{m} \times (150 - 20) = 100 \times 4.2 \times (20 - 11)$
$s_{m} = 0.582 J g^{- 1} K^{- 1}$

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Calorimetry

Standard IX Physics

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A mass of 50 g of a certain metal at 150∘C is immersed in 100 g of water at 11∘C. The final temperature is 20 ∘C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J g−1 K−1.

Heat lost by the metal = Heat gained by water ⟹mmSmΔTm=mwSwΔTw 50×sm×(150−20)=100×4.2×(20−11) sm=0.582 Jg−1K−1

A mass of $50 g$ of a certain metal at $150^{\circ} C$ is immersed in $100 g$ of water at $11^{\circ} C$ . The final temperature is $20^{\circ} C$ . Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$ .

Heat lost by the metal = Heat gained by water
$⟹ m_{m} S_{m} Δ T_{m} = m_{w} S_{w} Δ T_{w}$
$50 \times s_{m} \times (150 - 20) = 100 \times 4.2 \times (20 - 11)$
$s_{m} = 0.582 J g^{- 1} K^{- 1}$