Question

What is the final temperature when 71.8 g of water at 78.8°C is mixed with 33.6 g of water at 29.0°C? (The specific heat of water is 4.184 J/g·°C.)

Answer 1

Answer:

62.92°

Explanation:

given,

initial mass of water, m₁ = 71.8 g

initial temperature, T₁ = 78.8°C

another mass of water, m₂ = 33.6 g

another temperature of water, T₂ = 29° C

Final temperature of mix = ?

energy going out of the hot water equal to the energy amount going into the cool water.

$q_{lost} =q_{gain}$

$m_1 c \Delta T = m_1 c \Delta T$

$71.8 (78.8-x) = 33.6\times (x - 29)$

$105.4 x = 6632.24$

x = 62.92°

Hence, the final temperature of the mix is equal to 62.92°