Question

A coffee-cup calorimeter contains 130.0 g of water at 25.3 ∘C . A 124.0-g block of copper metal is heated to 100.4 ∘C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g⋅K . The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.3 ∘C .

Part A

Determine the amount of heat, in J , lost by the copper block.

Part B Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g⋅K .

Answer 1

The amount of heat lost by copper block = m_Cu x C_Cu x T

Where;

m_Cu = Mass of copper = 124 g

C_Cu = Specific heat of copper = 0.385 J g^-1 K^-1

T = final temperature – Initial temperature

     = 30.3 ^oC – 100.4 ^oC

    = - 70.1 ^oC

    = - 70.1 K (change in temperature in ^oC = K)

So,

Q = m_Cu x C_Cu x T

   = (124 g) x (0.385 J g^-1⋅K^-1) x (- 70.1 K)

   = - 3346.57 J

PART B:

The amount of heat gained by water = m_water x C_water x T

Where;

m_water = Mass of water = 130 g

C_water = Specific heat of copper = 4.184 J g^-1 K^-1

T = final temperature – Initial temperature

     = 30.3 ^oC – 25.3 ^oC

    = 5 ^oC

    = 5 K (change in temperature in ^oC = K)

So,

Q = m_water x C_water x T

   = (130 g) x (4.184 J g^-1⋅K^-1) x (5 K)

   = 2719.60 J