A 1.4-kg block of ice is at a temperature of -10∘C. If 6200 J of thermal energy is added to the ice, what is its final temperature?
Well, it sounds like the ice is going to have a meltdown. But let's crunch some numbers and find out!
To solve this problem, we can use the specific heat capacity of ice, which is about 2,090 J/kg⋅°C. We can use the formula:
Q = mcΔT
Where:
Q = thermal energy in Joules (6200 J)
m = mass of the ice in kg (1.4 kg)
c = specific heat capacity of ice (2090 J/kg⋅°C)
ΔT = change in temperature (unknown)
Plugging in the values, we get:
6200 J = 1.4 kg * 2090 J/kg⋅°C * ΔT
Simplifying, we find:
ΔT = 6200 J / (1.4 kg * 2090 J/kg⋅°C)
ΔT ≈ 3.17 °C
So, the final temperature of the ice will be -10 °C + 3.17 °C, which is approximately -6.83 °C.
Keep an eye out for slippery situations, Ice Block! Stay cool! 😄
To find the final temperature of the ice after adding thermal energy, we can use the formula:
Q = mcΔT
Where Q is the thermal energy added, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
Since we are given the thermal energy added (Q = 6200 J), the mass of ice (m = 1.4 kg), and the specific heat capacity of ice (c = 2100 J/kg⋅°C), we can rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 6200 J / (1.4 kg × 2100 J/kg⋅°C)
Now we can calculate the change in temperature:
ΔT = 6200 J / 2940 J/°C
ΔT ≈ 2.10 °C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = -10°C + 2.10°C
Final temperature ≈ -7.90°C
Therefore, the final temperature of the ice after adding 6200 J of thermal energy is approximately -7.90°C.