Question

A 12 gram piece of Cu at 475 oC is placed in contact with a 15 gram piece of Cr at 265 oC.

Answer 1

Answer:

349.22°C

Explanation:

Let the final temperature of the two pieces of metal be x.

Now, the warmer metal which is C u reduces from 475°C to x. Thus Δt for C u is; Δt1 = 475 - x.

The cooler metal Cr increases in temperature from 265°C to x. Thus, it's change in temperature is Δt for Cr is; Δt2 = x - 265.

Now from conservation of energy, the amount of energy leaving the C u metal is equal to the amount of energy entering the Cr metal.

Thus;

q_lost = q_gain

Where;

q_lost = m1•c1•Δt1

q_gained = m2•c2•Δt2

Now, c1 & c2 are the specific heat capacity of C u and Cr respectively.

From online tables, c1 = 0.385 J/g°C and c2 = 0.46 J/g°C

We are given;

m1 = 12g and m2 = 15g

Thus;

12 × 0.385 × (475 - x) = 15 × 0.46 × (x - 265)

2194.5 - 4.62x = 6.9x - 1828.5

6.9x + 4.62x = 2194.5 + 1828.5

11.52x = 4023

x = 4023/11.52

x = 349.22°C