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