Question

a 5.0-kg piece of lead at a temperature of 600°c is placed in a lake whose temperature is 15°c. determine the entropy change of (a) the lead piece, (b) the lake, and (c) the universe.

Answer 1

To determine the entropy change of the lead piece, lake, and the universe, you can use the formula ΔS = mcΔT/T. The entropy change of the lead piece is approximately 692 J/°C, the entropy change of the lake is approximately 1394 J/°C, and the entropy change of the universe is approximately 2086 J/°C.

Step-by-step explanation:

To determine the entropy change of the lead piece, lake, and the universe, we need to use the formula:

ΔS = mcΔT/T

where ΔS is the entropy change, m is the mass of the object, c is the specific heat capacity, and ΔT is the temperature change.

(a) For the lead piece:

ΔS_lead = (5.0 kg)(130 J/kg°C)(600°C - 15°C)/(600°C)

ΔS_lead ≈ 692 J/°C

(b) For the lake:

ΔS_lake = (m_water)(c_water)(ΔT_water)/T_water + (m_ice)(c_ice)(ΔT_ice)/(T_water + T_ice)

ΔS_lake = (m_water)(4186 J/kg°C)(15°C - 0°C)/(15°C) + (150 g)(2060 J/kg°C)(15°C - 0°C)/(0°C + 15°C)

ΔS_lake ≈ 1394 J/°C

(c) For the universe:

ΔS_universe = ΔS_lead + ΔS_lake

ΔS_universe ≈ 692 J/°C + 1394 J/°C

ΔS_universe ≈ 2086 J/°C