in an exp. on thr sp. heat of a metal a 0.5 kg block of the metal 150 150 degree celsius is drop in a copper calorimeter of water equivalent 0.025 kg containing 150 cm cube of water at 27 degree Celsius the final temperature is 40 degrees celsius compute the special heat of the metal
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Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T1 = 150°C
Final temperature of the metal, T2 = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C:
150 × 1 = 150 g
Fall in the temperature of the metal:
ΔT = T1 – T2 = 150 – 40 = 110°C
Specific heat of water, Cw = 4.186 J/g/°K
Specific heat of the metal = C
Heat lost by the metal, θ = mCΔT … (i)
Rise in the temperature of the water and calorimeter system:
ΔT′’ = 40 – 27 = 13°C
Heat gained by the water and calorimeter system:
Δθ′′ = m1CwΔT’
= (M + m′) Cw ΔT’ … (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m’) Cw ΔT’
200 × C × 110 = (150 + 25) × 4.186 × 13
∴ C = (175 × 4.186 × 13) / (110 × 200) = 0.43 Jg-1K-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.
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Given below are observations on molar specific heats at room temperature of some common gases. Observations The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-11 Thermal Properties of Matter » « A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-11 Thermal Properties of Matter
Initial temperature of the metal, T1 = 150°C
Final temperature of the metal, T2 = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C:
150 × 1 = 150 g
Fall in the temperature of the metal:
ΔT = T1 – T2 = 150 – 40 = 110°C
Specific heat of water, Cw = 4.186 J/g/°K
Specific heat of the metal = C
Heat lost by the metal, θ = mCΔT … (i)
Rise in the temperature of the water and calorimeter system:
ΔT′’ = 40 – 27 = 13°C
Heat gained by the water and calorimeter system:
Δθ′′ = m1CwΔT’
= (M + m′) Cw ΔT’ … (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m’) Cw ΔT’
200 × C × 110 = (150 + 25) × 4.186 × 13
∴ C = (175 × 4.186 × 13) / (110 × 200) = 0.43 Jg-1K-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.
Prev quesNext ques
Given below are observations on molar specific heats at room temperature of some common gases. Observations The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-11 Thermal Properties of Matter » « A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-11 Thermal Properties of Matter
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