Question

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

Answer 1

Answer:

T2 = 135.1°C

Explanation:

Given data:

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