Can anyone help me with this question

## Question

100g of water at 20^{\circ}C is poured into a beaker containing 200g of water at 80^{\circ}C. What is the final temperature of the mixture?

100g air pada 20^{\circ}C dituangkan ke dalam bikar yang mengandungi 200g air pada 80^{\circ}C. Apakah suhu air campuran tersebut?

A 30^{\circ}C

B 50^{\circ}C

C 60^{\circ}C

D 70^{\circ}C

### Answer

Hi Isac,

For this question, they are testing your understanding of thermal equilibrium. The final temperature of the mixture is when thermal equilibrium is reached between the two masses of water, which means that the two masses of water have the same final temperature.

We know heat flows from hot area to cold area, so there is a **net flow of heat from the 200g water** (the hotter substance) **to the 100g water** (the colder substance)

**200g of water loses heat energy** and the **100g of water gains heat energy.**

And we know that: E=mc\triangle\theta, where:

E - heat energy

m - mass of substance

c - specific heat capacity of substance

\triangle\theta - change in temperature

(for \triangle\theta we usually do the larger temperature - the smaller temperature)

specific heat capacity of water is usually about 4.18Jkg^{-1}{^\circ} C^{-1}, but to solve this you don’t actually need it because the c will cancel off on both sides of the equation.

m_1c\triangle\theta_1=m_2c\triangle\theta_2

200c(80-x) = 100c(x-20)

2(80-x)=x-20

160-2x=x-20

180=3x

x=60

Therefore the answer is C. Hope this helps!