Question

A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final temperature of the water is 31.5 oC. What is the heat change of the metal in joules? The specific heat of water is 4.184 J/goC

Answer 1

Answer:

Heat change of the metal = -6799 J

Explanation:

Step 1: Data given

Mass of the metal = 20.0 grams

Mass of water = 250.0 grams

Initial temperature of water = 25.0 °C

Final temperature of water = 31.5 °C

Step 2: Calculate the heat change of the metal in joules?

Heat lost =  heat gained

Qlost = -Qgained

Q = m*C*ΔT

m(metal) * C(metal) * ΔT(metal) = -m(water) * C(water) * ΔT(water)

⇒with m(metal) = the mass of metal = 20.0 grams

⇒with C(metal) = the specific heat of meteal = ?

⇒with ΔT(metal) = the change of temperature of the metal = ?

⇒with m(water) = the mass of water = 250.0 grams

⇒with C(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 31.5 °C - 25.0 °C = 6.5 °C

The heat change of metal = heat change of water

Q = 250 * 4.184 * 6.5

Q = 6799 J

Heat change of water = 6799 J

Heat change of the metal = -6799 J

Answer 2

Answer:

$Q_{metal} = -6799\,J$

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

$Q_{metal} = - Q_{w}$

$Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})$

$Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)$

$Q_{metal} = -6799\,J$