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90.0 mL of H2O is initially at room temperature (22.0∘C). A chilled steel rod at 2.0∘C is placed in the water. If the final temperature of the system is "21.1" ∘C, what is the mass of the steel bar?Specific heat of water = 4.18 J/g⋅∘CSpecific heat of steel = 0.452 J/g⋅∘C

Respuesta :

Answer:

The steel bar weighs 39.014 grams

Explanation:

Heat lost by water = - heat gained by steel

m(water) = 90 grams

m(steel) = TO BE DETERMINED

specific heat of water = 4.18 J/g°C

specific heat of steel = 0.452 J/g°C

Initial temperature T1 water = 22°C

Final temperature T2 water = 21.1 °C

Initial temperature T1 steel = 2°C

Final temperature T2 steel = 21.1°C

Heat transfer = m*Cp*ΔT

with Cp = specific heat

with ΔT = Change of temperature = T2 - T1

m(water) * Cp(water) * ΔT(water) = m(steel) *Cp(steel) *ΔT(steel)

90g * 4.18 * ( 21.1 - 22) = -m(steel) * 0.452 * (21.1-2)

-338.58  = - m(steel)*8.6784

m(steel) = 39.014 grams

The steel bar weighs 39.014 grams

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