Question

A 1.45 kg metal head of a hammer strikes a solid rock with a velocity of 6.43 m/s. Assuming all the kinetic energy of the hammer is retained by the hammer head and converted into thermal energy, how much will the hammer head increase in temperature?


c head = 0.145 kcal / kg°C

Answer 1

The increase in temperature of the hammer head is determined as 0.034 ⁰C.

Kinetic energy of the hammer

The kinetic energy of the hammer is calculated as follows;

K.E = ¹/₂mv²

K.E =0.5 x 1.45 x 6.43²

K.E = 29.98 J

Increase in temperature

Q = mcΔθ

Δθ = Q/mc

where;

• c is specific heat capacity =  0.145 kcal / kg°C
• Q = K.E = 29.98 J = 0.0072 kcal

Δθ = (0.0072)/(1.45 x 0.145)

Δθ = 0.034 ⁰C

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Answer 2

Find Kinetic energy

$\\ \rm\Rrightarrow KE=\dfrac{1{2}mv^2$

$\\ \rm\Rrightarrow KE=\dfrac{1}{2}(1.45)(6.43)^2$

$\\ \rm\Rrightarrow KE=29.98J=0.0072kcal$

Now

$\\ \rm\Rrightarrow Q=mc\Delta T$

• We need ∆T

$\\ \rm\Rrightarrow 0.0072=1.45(0.145)T$

$\\ \rm\Rrightarrow \Delta T=0.03°C$