Copper weighing 0.5 kg is dipped in 500 g of water, where it cools down from 80 degrees to 17 grams.

Copper weighing 0.5 kg is dipped in 500 g of water, where it cools down from 80 degrees to 17 grams. Calculate how many degrees the water will heat up.

Problem data: m1 (mass of taken copper) = 0.5 kg; m2 (mass of water) = 500 g (in SI system m2 = 0.5 kg); t0 (initial copper temperature) = 80 ºС; t (final temperature) = 17 ºС.

Reference values: Cm (specific heat of copper) = 400 J / (kg * K); Sv (specific heat of water) = 4200 J / (kg * K).

We express the change in temperature of 500 g of water from the equality: Cm * m1 * (t – t0) = Sv * m2 * Δt, whence Δt = Cm * m1 * (t0 – t) / (Sv * m2).

Let’s calculate: Δt = 400 * 0.5 * (80 – 17) / (4200 * 0.5) = 6 ºС.