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
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 :)
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