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Asked Oct 15, 2019

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A 20-gram piece of copper (c = 0.385 J/g degrees celsius) at 115 degrees Celsius is placed in contact with a 12-gram sample of magnesium (c = 1.024 J/g degrees celsius) at 38 degrees celsius. What will the final temperature of the two metals be when they reach an equilibrium condition?

Step 1

The formula for specific heat can be written below in which q represents the heat energy in Joules, m represents the mass of the substance in kilogram, c represents the specific heat and ∆t represents the change in the temperature.

q = mc∆t …… (1)

The amount of heat lost by the copper metal(q_{c}) is equal to amount of heat gained by sample of magnesium(q_{m}). Thus, q_{c}=q_{m}. This is shown in equation (2) in which m_{c} is the mass of copper, c_{c }is the specific heat for copper, ∆t_{c} is the change in temperature of the copper, m_{m }is the mass of magnesium , c_{m} is the specific heat for magnesium and ∆t_{m }is the change in temperature of the magnesium.

m_{c}c_{c}∆t_{c}= m_{m}c_{m}∆t_{m } …… (2)

m_{c}c_{c }(t_{i} -t_{f} )= m_{m}c_{m}(t_{f} -t_{i} ) …… (3)

Step 2

Substitute 20 g for mc, 115○C for ∆ti, 12 g for mm, 1.024 J/g○C for cm and 38 ○C for ∆ti in equation (3) as shown in equation (4).

Further, solve the above e...

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