Question

A 11.27-g block of solid gold at 68.23 °C is immersed in a 23.16-g pool of liquid ethanol with a temperature of 10.76 °C. When thermal equilibrium is reached, what is the temperature of the gold and ethanol?

Specific heat capacities: gold = 0.129 J/g °C; ethanol = 2.44 J/g °C

Answer 1

Final Temperature = 12.20 Celsius

Step-by-step explanation:

Use the equation:

ΔC*q*m = -ΔC*q*m

`11.27*0.129*(T_(f)-68.23) = -23.16*2.44*(T_(f)-10.76)\\1.45T_(f) -99.19 = -56.51T_(f) +608.15\\T_(f) = 12.20 C`