Calculate q when 0.10 g of ice is cooled from 10 degrees C to -75 degrees C (C ice= 2.087 J/ g* K)
q=mcdeltaT
=.10g*2.087j/g*k*(273.15+10)-(-75+273.15)
=.10g*2.087j/g*k*(283.15k-198.15k)
=17.74J
Why did the ice go to therapy? It had a chilling experience!
Now, let's calculate the change in temperature of the ice.
ΔT = (final temperature - initial temperature)
= (-75°C - 10°C)
= -85°C
Next, we can calculate the amount of energy transferred by cooling down the ice.
q = (mass) x (specific heat capacity) x (ΔT)
Since we have 0.10 g of ice, the mass is 0.10 g.
q = (0.10 g) x (2.087 J/g*°C) x (-85°C)
Now, let's crunch some numbers!
q = -17.7 J
So, q, which represents the amount of energy transferred, is equal to -17.7 J. But remember, negative values just mean that energy is being taken out of the system.
To calculate the amount of heat transferred, we can use the formula:
q = m * c * ΔT
Where:
- q is the amount of heat transferred
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
Given:
- m = 0.10 g (mass of ice)
- c = 2.087 J/g*K (specific heat capacity of ice)
- ΔT = -75°C - 10°C = -85°C (change in temperature)
Now, we can plug in the values into the formula to find the amount of heat transferred:
q = 0.10 g * 2.087 J/g*K * -85°C
q = -1.775 J
Therefore, the amount of heat transferred when 0.10 g of ice is cooled from 10°C to -75°C is -1.775 Joules. The negative sign indicates that heat is being released from the ice.
To calculate the amount of heat (q) required to cool a given mass (m) of substance from one temperature (T1) to another temperature (T2), you can use the formula:
q = m * C * ΔT
Where:
q is the amount of heat transferred (in joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in J/g*K), and
ΔT is the change in temperature (T2 - T1) of the substance.
In this case, we have:
m = 0.10 g (mass of ice)
C = 2.087 J/g*K (specific heat capacity of ice)
T1 = 10°C (initial temperature of the ice)
T2 = -75°C (final temperature of the ice)
First, let's calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = -75°C - 10°C
ΔT = -85°C
Now, we can calculate q using the formula mentioned above:
q = m * C * ΔT
q = 0.10 g * 2.087 J/g*K * -85°C
To proceed with this calculation, we need to convert the temperature change from Celsius (°C) to Kelvin (K) because the specific heat capacity is given in J/g*K. The Kelvin scale is the same as the Celsius scale, but with an offset of 273.15.
ΔT(K) = ΔT(°C) + 273.15
ΔT(K) = -85°C + 273.15
ΔT(K) = 188.15 K
Now we can substitute this value into the formula:
q = 0.10 g * 2.087 J/g*K * 188.15 K
Calculating the expression:
q ≈ 3.914 J
Therefore, approximately 3.914 J of heat is required to cool 0.10 g of ice from 10°C to -75°C.