Question

A block of iron at 33.0°C has mass 2.30 kg. If 3.50×104 J of heat are transferred to the block, what is its resulting temperature?

Answer 1

66.8°C

Step-by-step explanation:

dQ = m*cp*dT where:

m = mass of block

cp = specific heat of iron

dT = temperature change

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dQ = 3.5 10^4 J

cp = 0.45 Kj / Kg = 450 J / Kg

3.5 10^4 = 2.3 * 450 * dt-----dt = 35000 / 450 * 2.3 = 33.8 °

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Final temperature = 33 + 33.8 = 66.8 °C