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

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1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//