When 150 G Of Steam Cools From 275°C To 250°C

4 When 150 G Of Steam Drops In Temperature From 2750 C To 2500 C

Cleaned assignment instructions: Calculate the heat energy released when 15.0 g of steam drops in temperature from 275.0 °C to 250.0 °C, given the specific heat capacity (Cp) of H₂O as 4.184 J/g°C. Determine the heat released during this cooling process.

Paper For Above instruction

The process of cooling steam involves the loss of thermal energy as temperature decreases. To quantify this heat transfer, the formula used is:

Q = m × Cp × ΔT

where:

• Q is the heat energy transferred (in Joules)
• m is the mass of the substance, in grams
• Cp is the specific heat capacity (J/g°C)
• ΔT is the change in temperature, in °C (final minus initial)

Given data:

• Mass of steam, m = 15.0 g
• Initial temperature, T_initial = 275.0 °C
• Final temperature, T_final = 250.0 °C
• Specific heat capacity of H₂O, Cp = 4.184 J/g°C

Calculating the temperature change:

ΔT = T_final - T_initial = 250.0 °C - 275.0 °C = -25.0 °C

The negative sign indicates heat loss, but for magnitude of energy released, we consider the absolute value.

Applying the formula:

Q = 15.0 g × 4.184 J/g°C × (-25.0 °C) = -1564 J

The negative sign signifies heat is released by the system. Therefore, approximately 1564 Joules of heat are liberated as the steam cools from 275.0 °C to 250.0 °C.

In practical applications, the energy released can be used for heating, cooking, or generating power, exemplifying the importance of thermal energy calculations in thermodynamics.

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