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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>
I'm not positive but if I'm reading the question right it would be the big bang sorry if I'm wrong
Answer:
The diameter of the hole increases
Explanation:
Metals expand and contract with temperature. Whenever metal is heated, it usually expands in relation to its thermal expansivity. This expansion leads to a slight increase in surface area.
Once the surface area of the metal changes, this means that the dimensions of the whole metal surface changed. As a result, the diameter of the hole drilled in the metal plate will change also. In our case, the diameter of the hole will increase.
