Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 13.7 °C are mixed. The specific heat of water is 4.184 J/g°C.

Answer 1

Answer:

The final temperature of the system is 42.46°C.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))$

where,

c = specific heat of water= $4.18J/g^oC$

$m_1$ = mass of water sample with 100 °C= 50.0 g

$m_2$ = mass of water sample with 13.7 °C= 100.0 g

$T_f$ = final temperature of system

$T_1$ = initial temperature of 50 g of water sample= $100^oC$

$T_2$ = initial temperature of 100 g of water =$13.7^oC$

Now put all the given values in the given formula, we get

$50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))$

$T_f=42.46^oC$

The final temperature of the system is 42.46°C.