91影视

The equation that governs the heat required or released for or from a system whenever there is a finite temperature change is given by,

$\mathit{Q}\mathbf{=}\mathit{m}\mathit{C}\mathbf{鈭�}\mathit{T}$ ...(i)

For phase change, the equation that governs the heat released or absorbed is,

$\mathit{Q}\mathbf{=}\mathbf{-}\mathit{m}{\mathit{L}}_{\mathbf{v}}\mathbf{.}$

Here, the transition is from vapor to liquid state.

In case of water,

$$\begin{gathered} c=4190J/kg.k \\ {L}_v=2256.{10}^{3}J/kg \end{gathered}$$

For the water system, the heat supplied is,

$Q_w=m_w{c}_w鈭�T$

For the steam system,, the heat required is for raising the temperature and phase transition is,

$Q_s=m_s{c}_s鈭�T+m_s{L}_v$

These values are equal as the heat energy supply is equal,

$$\begin{gathered} m_w{c}_w鈭�T=m_s{c}_s鈭�T+m_s{L}_v \\ =m_s\left(c_s鈭�T+L_v\right) \end{gathered}$$ 鈥�(颈颈颈)

Change in temperature of water,

$$\begin{gathered} 鈭�T_w=\left(70.0K-28.0K\right) \\ =42.0K \end{gathered}$$

Change in temperature of steam, $鈭�T_s=65.0K$

From (iii), Rearrange for the ratio of masses,

$$\begin{gathered} \frac{m_s}{m_w}=\frac{c_w鈭�T}{c_w鈭�T_s+L_v} \\ =\frac{\left(4190J/kg脳K\right)\left(42.0K\right)}{\left(4190J/kg脳K\right)\left(65脳0K\right)+2256脳{10}^{3}J/kg} \\ =0.0696kg \end{gathered}$$

Thus, 0.0696 kg steam supplies equal heat as 1.00 kg of hot water.