If 1.9 kJ of heat is transferred to 96 g aluminum at 113°C, what would the

new temperature of the aluminum be? (The specific heat capacity of

aluminum is 0.897 J/g °C.)

T2 = 135.1°C

Explanation:

Given dа ta:

Mass of water = 96 g

Initial temperature = 113°C

Final temperature = ?

Amount of energy transfer = 1.9 Kj (1.9*1000 = 1900 j)

Specific heat capacity of aluminium = 0.897 j/g.°C

Solution:

Formula:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Now we will put the values in formula.

Q = m. c. ΔT

1900 j = 96 g * 0.897 j/g.°C * T2 - 113°C

1900 j = 86.112 j/°C * T2 - 113°C

1900 j / 86.112 j/°C = T2 - 113°C

22.1°C + 113°C = T2

T2 = 135.1°C