A steel casting weighing 2 kg has an initial temperature of 500°c; 25 kg of water initially at 25°c is contained in a perfectly

Question

7 days ago

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A steel casting weighing 2 kg has an initial temperature of 500°c; 25 kg of water initially at 25°c is contained in a perfectly

insulated steel tank weighing 5 kg. the casting is immersed in the water and the system is allowed to come to equilibrium. what is its final temperature? ignore the effects of expansion or contraction, and assume constant specific heats of 4.18 kj⋅kg−1⋅°c−1 for water and 0.50 kj⋅kg−1⋅°c−1 for steel.

Physics

2 answers:

Ostrovityanka [942] · Answer 1 · 2026-02-26T15:49:45+03:00

Ostrovityanka [942]7 days ago

5 0

Result:

26.6C

Rationale:

Employing a balance of energy:

The drop in internal energy of the casting must result in an increase in the internal energy of both the tank and the water, assuming that there is no heat loss from the tank as it is perfectly insulated.

$m_{casting}*C_{p,casting}*(T_{casting} - T_{final})= m_{water}*C_{p,water}*(T_{final} - T_{water}) + m_{tank}*C_{p,tank}*(T_{final} - T_{tank})$

$2*0.5*(500 - T_{final})= 70*4.18*(T_{final} - 25) + 5*0.5*(T_{final} - 25)\\\\(1+2.5+292.6)*T_{final} = 500+7315+62.5\\\\T_{final} = 26.6C$

kicyunya [1K] · Answer 2 · 2026-02-26T15:47:02+03:00

<span> The strategy involves using Q = m · c · ΔT three times. The temperature change for the hot casting is ΔT_hot = 500°C - Tf, and for the cold water and steel tank, it’s ΔT_cold = Tf - 25°C. We equate the heat lost from the hot casting during cooling to the heat gained by the cold tank during heating.
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, noting that K° = C° (for ΔT's).</span>