Question

31.0 g of copper pellets are removed from a 300° C oven and immediately dropped into 80.0 mL of water at 18.0 ° C in an insulated cup. ▼ Part A What will the new water temperature be? You may want to review (Pages 530-531) Express your answer with the appropriate units. Value Units Submit Request Answer

Answer 1

Solution)

Given, mass of copper =31 g

T1 =300 degrees , final temperature, T2= ?

We know, Specific heat of copper = 0.385 J/goC

Specific heat of water = 4.186 J/kg.oC

Given , volume V =80 mL= 80x10^-6 m^3

rho= 1000 kg/m^3

Now,

mw =rho*V =1000x80x10-6 = 0.08 kg

We know acc to Conservation of Energy,

Loss of energy = gain in energy

So, m*c (DeltaT) of copper =m*c(DeltaT) of water

(31)(0.385) [T -300] = - (0.08)(4186)(T- 18)

0.0356 [T -300] = (18- T)

0.0356 T - 10.69 = 18-T

T = 27.7 degrees

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Good luck!:)