If a special type of ice has a specific heat capacity of 2060 J/kg K. How much heat must be absorbed by 2.6 kg of ice at -27.00 C to raise it up to
0.00 C, before any melting takes place?

A. 140,000 J
B. 290,000 J
C. 56,000 J
D. None of the above

Q = 144612 Joules.

Explanation:

Given the following data;

Mass = 2.6 kg

Initial temperature = -27°C to Kelvin = 273 + (-27) = 246K

Final temperature = 0°C to Kelvin = 273K

Specific heat capacity = 2060 J/kgK.

To find the quantity of heat absorbed;

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 = 273 - 246

dt = 27 K

Substituting the values into the equation, we have;

Q = 144612 Joules.

we need to know the time interval during which the force acts. impulse of force = momentum change

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min