Question

A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calculate the specific heat of aluminum.

Answer 1

Answer:

Explanation:

Specific heat is defined as the amount of energy required to heat or cool 1 kg of a substance by 1°C.

It is given:

energy is 1080 joules,

mass is 140g; i.e. 0.14kg,

temperature change is 98.4 - 62.2 = 36.2°C

So the specific heat of aluminum = 1080 / 0.14 / 36.2

= 213.1 J/kg/°C

Answer 2

Answer:

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is $(\frac {J}{Kg.K})$

C = Q/m. ∆T

C – Specific heat $(\frac {J}{Kg.K})$

Q- heat energy (J)

M – Mass (Kg)

∆T- change in temperature (K)

Explanation:

Given data:

M= 140 g = 0.14 Kg

Q – 1080 Joules.

∆T – 98.4 – 62.2 = 36.2

Substituting  the given data in Equation

Specific heat of Aluminium  = $\frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})$