A block of iron at 39.0°C has mass 2.30 kg. If 3.50×104 J of heat are transferred to the block, what is its resulting temperature?
The specific heat of iron is 449 (J/kg*K)
Someone else had this question on here but I cannot get the right answer from the explanation given. Can someone please work this out and show me what my answer should be so I don't get my calculations all jumbled anymore?! Thank you :)
Q= m•c•ΔT
ΔT = Q/m•c=3.5•10^4/2.3•449=33.9º
T2=T1+ΔT= 39+33.9=72.9ºC=345.9 K
To find the resulting temperature of the iron block, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat transferred to the block (in Joules)
m = mass of the iron block (in kilograms)
c = specific heat of iron (in J/kg*K)
ΔT = change in temperature of the block (final temperature - initial temperature)
In this case, the heat transferred (Q) is given as 3.50×10^4 J, the mass (m) is 2.30 kg, and the specific heat (c) is 449 J/kg*K.
We need to solve for ΔT, so rearrange the formula:
ΔT = Q / (m * c)
Now, substitute the given values into the equation:
ΔT = (3.50×10^4 J) / (2.30 kg * 449 J/kg*K)
Calculating this expression will give us the change in temperature (ΔT). Let's solve it:
ΔT ≈ 30.27 K
To find the resulting temperature, add the change in temperature to the initial temperature:
Resulting temperature = initial temperature + ΔT
Resulting temperature = 39.0°C + 30.27 K
It's important to convert the initial temperature from Celsius to Kelvin by adding 273.15:
Resulting temperature ≈ 39.0°C + 30.27 K + 273.15
Resulting temperature ≈ 342.42 K
So, the resulting temperature of the iron block is approximately 342.42 Kelvin.