Answer:
Final temperature of water would have been equal to 54.5 °C.
Step-by-step explanation:
Given:
Amount of transferred energy = 65 K J
Mass of water = 450 g
Initial temperature = 20°C
To find the final temperature of water:
Formula for Heat capacity is given by
Q = m×c×Δt ........................................(1)
where:
Q = Heat capacity of the substance (in J)
m=mass of the substance being heated in grams(g)
c = the specific heat of the substance in J/(g.°C)
Δt = Change in temperature (in °C)
Δt = (Final temperature - Initial temperature) = T(f) - T(i) .......................(2)
Q = 65 K J,
As Q is given in K J, to convert kilo-joules into joules, 1 K J = 1000 J
Therefore 65 K J = 65,000 J
Q = 65,000 J
Mass of water = m = 450 g
Initial temperature of water = T(i) = 20°C
Specific heat of water is c = 4.186 J /g. °C
Substituting these in equation (1), we get
Q = m×c×Δt
Rearranging the terms for Δt,
Δt = Q/(m×c)
Δt = 65,000/(450×4.186) = 65,000/1883.7
Δt = 34.5 °C ........................................(3)
Substituting (3) in (2) and solving for T(f):
Δt = T(final) - T(initial) = T(f)- T(i)
Rearranging the equation for T(f),
T(f) = Δt + T(i)
T(f) = 34.5 + 20
T(f) = 54.5 °C
Final temperature of water would have been equal to T(f) = 54.5 °C.