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Home » 45 g of water at 50o C in a beaker is cooled when 50 g of copper at 18oC is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g-1 K-1 and that of water is 4.2 J g-1 K-1. State the assumptions used.
45 g of water at 50o C in a beaker is cooled when 50 g of copper at 18oC is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g-1 K-1 and that of water is 4.2 J g-1 K-1. State the assumptions used.
Calorimetry | Class 10 | ICSE | Physics | Selina
45 g of water at 50o C in a beaker is cooled when 50 g of copper at 18oC is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g-1 K-1 and that of water is 4.2 J g-1 K-1. State the assumptions used.

Solution:

According to the question,

Mass of water (m1) is 45 g

Temperature of water (T1) is 500 C

Mass of copper (m2) is 50 g

Temperature of copper (T2) is 180 C

We have to find the final temperature (T)

Also, the specific heat capacity of the copper c2 is 0.39 J / g / K

And the specific heat capacity of water c1 is 4.2 J / g / K

Using the expression for heat energy, Q = mc(change in temperature), we can write expressions for –

Heat lost by water = m1c1 (T1 – T)

And heat gained by copper = m2c2 (T – T2)

And as we know that when there is no heat loss to surroundings, the heat gain is equal to the heat loss. Therefore, we have –

m1c1 (T1 – T) = m2c2 (T – T2)

Upon re-arranging and solving for temperature T, we get => T = (m1c1T1 + m2c2T2) / (m2c2 + m1c1)

Putting known values, =>T = (45 × 4.2 × 50) + (50 × 0.39 × 18) / (45 × 4.2) + (50 × 0.39)

T = (9450 + 351) / (189 + 19.5) = 9801 / 208.5

Therefore, T = 470 C

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