Question

What would the expected temperature change be (in °F) if a 0.5 gram sample of water released 50.1 J of heat energy? The specific heat of liquid water is 4.184 J/g-°C

Answer 1

Given,

Mass of water sample = 0.5 g

Heat energy(q) = 50.1 J

specific heat of liquid water(C) = 4.184 J/ g ^oC

Now, we know the formula,

q = m x C x T

Here, T is the temperature change.

Rearranging the above formula,

T = q / (m x C)

T = 50.1 J / (0.5 g x 4.184 J/g ^oC)

T = 23.948 ^oC

Now, converting ^oC to ^oF,

T(^oF) = [(9/5) x T(^oC)] + 32

Substituting the values,

T(^oF) = (9/5) x 23.948 + 32

T(^oF) = 75.1 ^oF Or 75 ^oF (2 S.F)