Question

If a special type of ice has a specific heat capacity of 2060 J/kg K. How

Answer 1

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;

$Q = mcdt$

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;

$Q = 2.6*2060*27$

Q = 144612 Joules.