Question

How much energy is required to changea 48 g ice cube from ice at -14 °C tosteam at 118 °C? The specific heat of iceis 2090 J/kg . °C, the specific heat of wa-ter is 4186 J/kg . ° C, the specific heat ofstream is 2010 J/kg . ° C, the heat of fusionis 3.33 x 10^5 J/kg, and the heat of vaporiza-tion is 2.26 x 10^6 J/kg.Answer in units of J.

Answer 1

Given data

*The given mass is m = 48 g = 48 × 10^-3 kg

*The given initial temperature of the ice is T = -14 °C

*The given temperature of the steam is t = 118 °C

*The specific heat of ice is c = 2090 J/kg °C

*The specific heat of water is s = 4186 J/kg °C

*The specific heat of the stream is 2010 J/kg ° C

*The heat of vaporization is 2.26 x 10^6 J/kg

*The heat of fusion is 3.33 x 10^5 J/kg

The formula for the energy is required to change a 48 g ice cube from ice at -14 °C to

steam at 118 °C is given as

$E=mc\Delta T+mL_f+ms\Delta t_1+mL_v+ms_s\Delta t_2$

Substitute the values in the above expression as

$\begin{gathered} E=(48\times10^{-3}\text{)}\lbrack(2090\times14)+(3.33\times10^5)+(4186\times104)+(2.26\times10^6)+(2010\times14) \\ =148115.7\text{ J} \end{gathered}$