Question

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.

Answer 1

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)

Answer 2

26.6C

Step-by-step explanation:

Using an energy balance:

Decrease in internal energy of casting must lead to an increase in internal energy of the tank and water, assuming that no heat flows out of the tank - 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`