Question

What is the specific heat capacity of a 250g object that needs 34.125kJ to have its own temperature raised by 65.0∘C

Answer 1

Answer:

Specific heat capacity, = 2.1J/g°C

Explanation:

Given the following data;

Mass = 250g

Quantity of heat = 34.125kJ = 34.125 * 1000 = 34125 Joules.

Change in temperature = 65.0∘C

To find the specific heat capacity;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making c the subject of formula, we have;

$c = \frac {Q}{mdt}$

Substituting into the equation, we have;

$c = \frac {34125}{250*65}$

$c = \frac {34125}{16250}$

Specific heat capacity, = 2.1J/g°C