A medium-sized pear provides about 102 Calories of energy. HINT (a) Convert 102 Cal to joules. J (b) Suppose that amount of energy is transformed into kinetic energy of a 2.03 kg object initially at rest. Calculate the final speed of the object (in m/s). m/s (c) If that same amount of energy is added to 3.79 kg (about 1 gal) of water at 23.7°C, what is the water's final temperature (in °C) ? The specific heat of water is c = 4186 J (kg · °C). °C

(a) 426.8 J

(b) 20.5 m/s

(c) 23.727 °C

Explanation:

(a)

E = Energy of medium-sized pear = 102 cal

we know that, 1 cal = 4.184 J

hence

E = 102 (4.184 J)

E = 426.8 J

(b)

KE = kinetic energy of the object = E = 426.8 J

m = mass of the object = 2.03 kg

v = speed of the object = ?

Kinetic energy of the object is given as

KE = (0.5) m v²

inserting the values

426.8 = (0.5) (2.03) v²

v = 20.5 m/s

(c)

Q = Amount of heat added to water = E = 426.8 J

m = mass of water = 3.79 kg

c = specific heat of water = 4186 J / (Kg °C)

T₀ = initial temperature = 23.7 °C

T = Final temperature = ?

Using the equation

Q = m c (T - T₀)

426.8 = (3.79) (4186) (T - 23.7)

T = 23.727 °C