Question

200g steam at 100° C is introduced on 800g ice at 0°C.Find the final temperature of the mixture​

Answer 1

The final temperature of the mixture​ is 20°C.

Explanation:

Given that,

Mass of steam = 200 g

Mass of ice = 800 g

Initial temperature = 100°C

Initial temperature = 0°C

We need to calculate the final temperature

Using formula of heat

For steam,

$Q=m_{steam}c\Delta T$...(I)

For ice,

$Q=m_{ice}c\Delta T$.....(II)

From equation (I) and (II)

$m_{steam}c\Delta T=-m_{ice}c\Delta T$

Negative sign shows the heat loss

Put the value into the formula

$200\times2.08(T_{f}-100)=-800\times2.10(T_{f}-0)$

$416T_{f}-41600=-1680T_{f}$

$T_{f}=\dfrac{41600}{2096}$

$T_{f}=20^{\circ}C$

Hence, The final temperature of the mixture​ is 20°C.

Learn more :

Topic : Heat loss

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