The specific heat capacity of water is 4.186 J/g °C Calculate the temperature change when 8000 J of heat is added to 3g of water.
Use the equation q=mcT

Explanation:

Energy (Q) J

Change of temperature (ΔT) °C

Mass (m) kg

Substance (optional) Custom ▾

Specific heat capacity (c) J/(kg·K)

HOPE IT HELPS

Solution given:

specific heat capacity of water [c]4.186 J/g °C

temperature[∆T]=?

mass[m]=3g

heat[Q]=8000J

we have

Q=mc∆T

8000=3*4.186*∆T

∆T=8000/12.558

∆T=637.04°C

the temperature change is 637.04°C.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

the answer is 75.0%.