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A 150 g piece of iron (Cp = 25.09 J/(mol-°C)) was heated to a temperature of
47°C and then placed in contact with a 275 g piece of copper at 20°C (Cp = 25.46
J/(mol-°C)). What is the final temperature of the two pieces of metal?

Respuesta :

Final temperature of the two pieces of metal is 32.25°C

Temprature is the degree of hotness or coldness of the object

Here given data is

Mass of iron = 150 g

Mass of copper = 275 g

Specific heat of iron = 25.09 J/mol°C

Specific heat of copper = 25.46J/mol°C

Temprature in iron = 47°C = initial temprature = T₁

Temprature in copper = 20°C = initial temprature = T₁'

So here we have to find final temprature of  two pieces of metal =?

So, the formula is Q =mcΔT

Q =  (150 g - 275 g) × (25.09 J/mol°C - 25.46J/mol°C) × ( 47°C-20°C )

Q = 125×0.37×27

Q = 243 J

243 J/125×0.37 = 5.25°C = change in temprature

Initial temprature  = 27°C

Final temprature = ?

ΔT = T₂ - T₁

27°C = T₂ - 5.25°C

T₂ = 27°C +  5.25°C

T₂ = 32.25°C

Final temperature of the two pieces of metal is  32.25°C

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