Question

Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water are Lf = 3.33 ✕ 105 J/kg and c = 4186 J (kg · °C) . HINT (a) Calculate the energy (in J) required to melt all the ice into liquid water. (Enter your answer to at least three significant figures.) J (b) How much energy (in J) remains to raise the temperature of the liquid water? (Enter your answer to at least three significant figures.) J (c) Determine the final temperature of the liquid water in Celsius. °C

Answer 1

Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

(c) 45 °C

Explanation:

(a)

$L_{f}$ = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

$Q_{f}$ = Energy required to melt the ice

Energy required to melt the ice is given as

$Q_{f}$ = m $L_{f}$

$Q_{f}$ = (1.63) (3.33 x 10⁵)

$Q_{f}$ = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q  = Amount of energy remaining to raise the temperature

Using conservation of energy

E = $Q_{f}$ + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

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

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

T = 45 °C