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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Answer:
8977.7 kg/m^3
Explanation:
Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3
Reading of balance when block is immersed in water = 4.3 N
According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.
Buoyant force = weight of the water displaced by the block
Buoyant force = Volume of water displaced x density of water x g
= 55 x 10^-6 x 1000 x .8 = 0.539 N
True weight of the body = Weight of body in water + buoyant force
m g = 4.3 + 0.539 = 4.839
m = 0.4937 kg
Density of block = mass of block / volume of block
=
Density of block = 8977.7 kg/m^3