IDNLearn.com provides a platform for sharing and gaining valuable knowledge. Discover prompt and accurate answers from our experts, ensuring you get the information you need quickly.

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


Sagot :

Answer:

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;

[tex] Q = mcdt[/tex]

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;

[tex] Q = 2.6*2060*27[/tex]

Q = 144612 Joules.

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.