Question

A 500-g aluminum container holds 300 g of water. The water and aluminum are initially at 40∘C. A 200-g iron block at 0∘C is added to the water. Assume the specific heat of iron is 450 J/kg⋅∘C, the specific heat of water 4180 J/kg⋅∘C and the specific heat of aluminum is 900 J/kg⋅∘C

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1Determine the final equilibrium temperature.    2.Determine the change in thermal energy of the aluminum

3.Determine the change in thermal energy of the water. 4. Determine the change in thermal energy of the iron.

show work please

Answer 1

1.

m_a = mass of aluminium = 500 g = 0.5 kg

m_w = mass of water = 300 g = 0.3 kg

m_i = mass of iron = 200 g = 0.2 kg

T_awi = initial temperature of aluminium and water = 40

T_ii = initial temperature of iron = 0

T = final equilibrium temperature

c_i = specific heat of iron = 450

c_w = specific heat of water = 4180

c_a = specific heat of aluminium = 900

using conservation of heat

Heat lost by aluminium and water = Heat gained by iron

m_w c_w (T_awi - T) + m_a c_a (T_awi - T) = m_i c_i (T - T_ii)

(0.3) (4180) (40 - T) + (0.5) (900) (40 - T) = (0.2) (450) (T - 0)

T = 38 C

2.

Q_a = change in thermal energy of aluminium = m_a c_a (T_awi - T) = (0.5) (900) (40 - 38) = 900 J

3.

Q_w = change in thermal energy of water = m_w c_w (T_awi - T) = (0.3) (4180) (40 - 38) = 2508 J

4.

Q_i = change in thermal energy of iron = m_i c_i (T - T_ii ) = (0.2) (450) (38 - 0) = 3420 J