Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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 :

Lanuel

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.

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.