Question

10 10 gram of ice at zero degree centigrade is mixed with hundred gram of water at 50 degree centigrade what is the resultant temperature of the mixture

Answer 1

HEY BRO HERE IS YOUR ANSWER

10 g of  ice + 100 g of water = 110 g

100 g of water =50°c

1 g of water = 50°c/100 =0.5°c

therefore for 110 g of solution temperature will be

110°c×0.5°c=55°c

so bro i hope you have got your answer

so mark me as Brainly.

Answer 2

The resultant temperature of the system is, $47.6^0C$

Explanation:

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

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.11J/g.^oC$

$c_2$ = heat capacity of water = $4.18J/g.^oC$

$m_1$ = mass of ice = 10 g

$m_2$  = mass of water = 100 g

$T_f$= final temperature of system = ?

$T_1$= initial temperature of ice = $0.0^oC$

$T_2$= initial temperature of water = $50^oC$

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

$10g\times 2.11J/g.^oC\times (T_f-0.0)^oC=-100g\times 4.18J/g.^oC\times (T_f-50)^oC$

$T_f=47.6^0C$

Therefore, the resultant temperature of the system is, $47.6^0C$

Learn more more about resultant temperature

https://brainly.com/question/13823212

https://brainly.com/question/13294103