Question

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that the pan is placed on an insulated pad. What would be the final temperature (in °C) of the pan and water if 0.0300 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan?
Cw= 4186, Cal=900, Lvapor=2256000
Find final temperature

Answer 1

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

• mass of water, $m_w=0.25\ kg$

• initial temperature of water, $T_i_w=20^{\circ}C$

• initial temperature of pan, $T_i_p=173^{\circ}C$

• mass of pan, $m_p=0.6\ kg$

• mass of water evapourated, $m_v=0.03\ kg$

• specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

• specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

• latent heat of vapourization, $L=2256000\ J.kg^{-1}$

Using the equation of heat:

Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$