# What is the final temperature of the copper and water given that the specific heat of copper is 0.385 J/(g * "^oC)?

## A hot lump of 48.7 g of copper at an initial temperature of 76.8 C is placed in 50.0 mL of ${H}_{2} O$ initially at 25.0°C and allowed to reach thermal equilibrium. Assume no heat is lost to surroundings.

Jul 15, 2017

T_f = 29.3 °C

#### Explanation:

Here, we're setting the heat values derived given the equation:

$q = m {C}_{s} \Delta T$

equal to each other, as such,

${q}_{w a t e r} + q \left(C u\right) = 0$
$\therefore {q}_{w a t e r} = - q \left(C u\right)$

Before we start, I will assume the density of water is $\text{1.00 g/mL}$.

50.0 g*(4.184J)/(g*°C)*(T_f-25.0°C) = -[48.7g*(0.385J)/(g*°C)*(T_f-76.8°C)]

209.2T_f - 5230 °C = -18.75T_f + 1440 °C

227.95T_f = 6670 °C

therefore T_f = 29.3°C