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

To find the final temperature when the water and pan reach thermal equilibrium, we need to calculate the heat lost by the pan and the heat gained by the water. By setting these two equal to each other and solving for the final temperature, we can find the answer.

Step-by-step explanation:

In this problem, we need to find the final temperature when the water and pan reach thermal equilibrium. First, we need to calculate the heat lost by the pan and the heat gained by the water.

We can use the formula: Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Answer 2

`T_f=5.0116^(\circ)C`

Step-by-step 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* 900* (173-T_f)=0.03* 2256000+(0.25-0.03)* 4186* (T_f-20)`

`T_f=5.0116^(\circ)C`