Engineering, 15.02.2020 02:27 10040813

An open container holds 0.550 kg of ice at -15.0 C. The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 800.0 J>min for 500.0 min.

(a) After how many minutes does the ice start to melt?
(b) After how many minutes, from the time when the heating is first started, does the temperature begin to rise above 0.0 C?
(c) Plot a curve showing the temperature as a function of the elapsed time. 17.47. CP What must the initial speed of a lead?

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Answer from: JesuGranger

a) t = 21.72 minutes

b) Total Time = 255.47 mins

Explanation:

Given:

- Mass of ice m = 0.55 kg

- Initial Temperature T_i = -15 C

- Rate of Heat in-put W_in = 800 J /min

- Specific Heat capacity of ice c_p = 2106 J/kgK

Find:

(a) After how many minutes does the ice start to melt?

(b) After how many minutes, from the time when the heating is first started, does the temperature begin to rise above 0.0 C?

Solution:

- The amount of heat required by a mass m of ice to raise its temperature from initial to T is given by the expression:

Q_in = m*c_p* ( T - T_i )

- The melting point of ice is T = 0 C

- The amount of Heat input is given by:

Q_in = W_in*t

Where, t is time in minutes

- Combine the two expressions:

t = m*c_p* ( T - T_i ) / W_in

- plug in the values to compute t:

t_1 = 0.55*2106*(0 - (-15)) / 800

t_1 = 21.72 minutes

- The amount of time it takes to raise the temperature of ice to its melting point is t = 21.72 mins.

- As soon as the ice reaches its melting point, For the temperature to rise again, it must break the inter-molecular forces between molecules and cause a phase transformation from solid to liquid. So the amount of heat required for ice to transform into water is called the latent heat of fusion. Hence,

W_in*t = m*L_f

t_2 = m*L_f / W_in

Plug values in: t_2 = 0.55*(3.4*10^5) / 800

t_2 = 233.75 mins

- Hence, the total time taken from initial temperature to the point it starts rising from 0 C is:

Total Time = t_1 + t_2

Total Time = 21.72+ 233.75

Total Time = 255.47 mins

An open container holds 0.550 kg of ice at -15.0 C. The mass of the container can be ignored. Heat i

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An open container holds 0.550 kg of ice at -15.0 C. The mass of the container can be ignored. Heat i...