Question

4. What quantity of heat is required to raise the temperature of 100

grams of water from 25°C to 45°C? I

Answer 1

Q = 836.4 Joules.

Step-by-step explanation:

Given the following data;

Mass = 100 grams

Initial temperature = 25°C

Final temperature = 45°C

We know that the specific heat capacity of water is equal to 4.182 J/g°C.

To find the quantity of heat;

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.

dt = T2 - T1

dt = 45 - 25

dt = 20°C

Substituting the values into the equation, we have;

`Q = 100*4.182*20`

Q = 836.4 Joules.