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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>
1) In first case, both the objects(hand & tea) are in direct contact, so heat will flow by the process of "Conduction"
2) In this case, objects are not in direct contact, and heat is transferring from one to another by "Convection"
3) Here, there is no contact between the objects (our body & sun), sun light with heat comes through the process of "Radiation"
In short, Your Answers would be:
1) Conduction
2) Convection
3) Radiation
Hope this helps!
Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
