Question

A 100.0g sample of nickel is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 25.0 oC. After the metal cools, the final temperature of the metal and the water is 29.8 oC. Calculate the specific heat capacity of nickel from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. Specific heat of water

Answer 1

Answer:

0.429 J/g.°C

Explanation:

Step 1: Calculate the heat absorbed by the water (Qw)

We will use the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in the temperature

Qw = 4.184 J/g.°C × 150. g × (29.8°C-25.0°C) = 3012 J

Step 2: Calculate the heat released by the sample of nickel

According to the law of conservation of energy, the sum of the heat lost by the sample of nickel and the heat absorbed by the water is zero.

Qw + QNi = 0

QNi = -Qw = -3012 J

Step 3: Calculate the specific heat capacity of nickel

We will use the following expression.

QNi = c × m × ΔT

c = QNi/m × ΔT

c = -3012 J/100.0 g × (29.8°C-100.0°C) = 0.429 J/g.°C