Suppose during a lab that 65kJ of energy were transferred to 450 g of water at 20 degree Celsius. What would have been the final temperature of the water

Final temperature of water would have been equal to 54.5 °C.

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.