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