Answer:
Explanation:
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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
Thus, we insert mass, specific heat and temperatures to obtain:
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
Now, we plug in to obtain:
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
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Answer:
m = 180 g
Explanation:
Given data:
Energy absorbed = 108 J
Mas of gold = ?
Initial temperature = 25°C
Final temperature = 29.7 °C
Specific heat capacity of gold = 0.128 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT =29.7 °C - 25°C
ΔT = 4.7 °C
108 J = m ×0.128 J/g.°C ×4.7 °C
108 J = m ×0.60 J/g
m = 108 J/0.60 J/g
m = 180 g
Heat energy is absorbed by the substance
Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice 
= latent heat of ice
Latent heat of ice = 334 KJ
⇒ = 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.
