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

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;

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;

Substituting into the equation, we have;

Specific heat capacity, = 2.1J/g°C

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

the answer is a. at the equator