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Plan: Use Q = m · c · ΔT three times. Hot casting cools ΔT_hot = 500°C -
Tf. Cold water and steel tank heat ΔT_cold = Tf - 25°C. Set Q from hot
casting cooling = Q from cold tank heating.
here
m_cast · c_steel · ΔT_hot = (m_tank · c_steel + m_water · c_water) · ΔT_cold
m_cast · c_steel · (500°C - Tf) = (m_tank · c_steel + m_water · c_water) · (Tf - 25°C)
2.5 kg · 0.50 kJ/(kg K°) · (500°C - Tf) = (5 kg· 0.50 kJ/(kg K°) + 40 kg· 4.18 kJ/(kg K°)) · (Tf - 25°C)
Solve for Tf, remember that K° = C° (i.e. for ΔT's) </span>
Answer: The correct answer is : A, B, C, D. All statements are correct.
Explanation: This case is called non-competitive inhibition. As a result of the inhibition of the enzyme, there will eventually be less substrate that becomes the final product and this route will be closed. Enzyme 1 has two binding sites.
Apart from cutaneous respiration<span> present in all </span>species<span>, most lissamphibians are born in an aquatic larval stage with gills. After metamorphosis, they develop lungs to breathe on land. The larvae of urodeles and apods present external, filamentous and highly branched gills which allow them to breathe underwater.
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