A solid ball of mass 10 kg at 40-C is gently placed in a liquid of mass 20 kg at 20-C and of specific heat capacity 1 cal-g-C- When the thermal equilibrium is attained- temperature of the system is 35-C- The specific heat capacity of solid ball is- Neglect heat capacity of the vessel

Name: CENGA - Grade 11 - Physics - Thermal properties of matter - Q128
Uploaded: 2022-07-04T10:36:42+05:30
Description: Formula used: nbsp;Q=msΔθ Given: nbsp;m_1=10 kg, θ_1=40 ^∘C, m_2=20 kg, θ_2=20 ^∘C, θ=35 ^∘C, s_2=1 cal/g^∘C Heat gain by liquid =Heat loss by solid ...

Formula used: nbsp;Q=msΔθ Given: nbsp;m_1=10 kg, θ_1=40 ^∘C, m_2=20 kg, θ_2=20 ^∘C, θ=35 ^∘C, s_2=1 cal/g^∘C Heat gain by liquid =Heat loss by solid ...

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Byju's Answer

Standard VIII

Physics

Floating and Sinking

A solid ball ...

Question

A solid ball of mass $10 k g at 40^{\circ} C$ is gently placed in a liquid of mass $20 k g at 20^{\circ} C$ and of specific heat capacity $1 c a l / g^{\circ} C$ . When the thermal equilibrium is attained, temperature of the system is $35^{\circ} C$ . The specific heat capacity of solid ball is: (Neglect heat capacity of the vessel)

2 c a l / g^{\circ} C

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3 c a l / g^{\circ} C

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1 c a l / g^{\circ} C

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6 c a l / g^{\circ} C

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Solution

The correct option is D $6 c a l / g^{\circ} C$
Formula used: $Q = m s Δ θ$

Given: $m_{1} = 10 k g, θ_{1} = 40^{\circ} C, m_{2} = 20 k g, θ_{2} = 20^{\circ} C, θ = 35^{\circ} C, s_{2} = 1 c a l / g^{\circ} C$

Heat gain by liquid =Heat loss by solid ball.
$Q_{l o s s} = Q_{g a i n}$
$m_{1} s_{1} (40 - 35) = m_{2} s_{2} (35 - 20)$
$10 \times s_{1} \times 5 = 20 \times 1 \times 15$
$s_{1} = 6 c a l / g^{\circ} C$

Final Answer : $(d)$ .

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Floating and Sinking

Standard VIII Physics

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A solid ball of mass 10 kg at 40∘C is gently placed in a liquid of mass 20 kg at 20∘C and of specific heat capacity 1 cal/g∘C. When the thermal equilibrium is attained, temperature of the system is 35∘C. The specific heat capacity of solid ball is: (Neglect heat capacity of the vessel)

The correct option is D 6 cal/g∘CFormula used: Q=msΔθ Given: m1=10 kg, θ1=40 ∘C, m2=20 kg, θ2=20 ∘C, θ=35 ∘C, s2=1 cal/g∘C Heat gain by liquid =Heat loss by solid ball. Qloss=Qgain m1s1(40−35)=m2s2(35−20) 10×s1×5=20×1×15 s1=6 cal/g∘C Final Answer : (d).