Question

please help me. Thanks

A piece of copper metal weighing 36.3 g is initially at 100.0 degree C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20.0 degree C. After stirring, the final temperature of both copper and water is 25.0 degree C. Assuming no heat losses, an that the specific heat capacity of water is 4.184 J/g degreeC, what is the specific heat capacity of the copper in J/g degreeC?

Answer 1

Here a very important assumption which has been state in the question is that we have to assume there is no heat lost.

Which implies that  all the heat lost by the copper winds up in the water.

so it goes that

Q_copper =Q_water

we know that

heat gained or lost Q = ms$\Delta$T

where m is the mass in g

s is the specific heat capacity in J/g^oC

and $\Delta$T = heat lost or gained

By substitution, we have (copper values on the left, water values on the right):

(mass) (ΔT) (s_copper) = (mass) (Δt) (s_water)

Putting the numbers in place gives us:

(36.3 g) (100°C-25°C) (s_copper) = (50 g) (25°C-20°C) (4.184 J g¯¹ °C¯¹)

s_copper = (50 x 5 x 4.184)/(36.3 x 75)

s_copper = 0.384 J g¯¹ °C¯¹ This is the specific heat capacity of copper