4. What quantity of heat is required to raise the temperature of 100 grams of water from 25°C to 45°C? I

Q = 836.4 Joules.

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;

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 = 836.4 Joules.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)