Answer:
0.178 °C
Explanation:
Given:
The frictional force, F = 48 N
Number of rubs, n = 18
Distance covered per rub, d = 7.16 cm = 0.0716 m
Total distance covered in 18 rubs, D = 0.0716 × 18 = 1.288 m
mass of the tissues warmed, m = 0.1 kg
Now,
Work done, W = Force × Displacement
W = 48 × 1.288 = 61.8624 J
Now,
Work done is the change in heat
W = ΔQ = 61.8624 J
also,
ΔQ = mCΔT
where, C is the specific heat
for human tissue, C = 3470 J/kg°C
thus,
61.8624 J = 0.1 × 3470 × ΔT
or
ΔT = 0.178 °C
Hence, the change in temperature is 0.178 °C
The increase in temperature of the tissue is 0.178 ⁰C.
The work done by friction is calculated as follows;
[tex]W = Fd\\\\W = 48 \times (18 \times 0.0716)\\\\W = 61.86 \ J[/tex]
The increase in temperature of the tissue is calculated as follows;
[tex]Q = W\\\\Q = m c \Delta T[/tex]
where;
[tex]\Delta T = \frac{Q}{mc} \\\\\Delta T = \frac{61.86}{0.1 \times 3470} \\\\\Delta T = 0.178 \ ^0 C[/tex]
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