Question

A drilling machine is used to drill a hole in a 10kg metal block. Power of the drilling machine is 1 KW and it is used for 5 minutes. If 40% of the
work done by the drilling machine is converted into heat in the metal block, the rise in its temperature is (specific heat-
0.12cal gm-1°C-1 and J=4.2 J/cal)
(A) 20°C
(B) 25°C
(C) 35°C
(D) 50°C

Answer 1

Explanation:

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Answer 2

Answer:

The rise in temperature is 0.2646°C.

Explanation:

The rise in temperature is calculated as,

$\Delta T=\frac{Q}{ms}$         (1)

Where,

ΔT=rise in temperature

Q=required energy

m=mass of the block

s=specific heat capacity

From the question we have,

The mass of the block(m)=10kg

The power of the block(P)=1kW=1×10³ W

The time during which the drilling machine is used(t)=5minutes=300 seconds

The specific heat(s)=0.12cal gm⁻¹°C⁻¹=0.504 Joule gm⁻¹°C⁻¹

Now, first, let's calculate "Q".

$Q=\frac{P}{t}$    (2)

We get the following by inserting the numbers in equation (2):

$Q=\frac{1\times 10^3}{300}$

$Q=\frac{10}{3}$     (3)

If 40% of the work done by the drilling machine is converted into heat that means 40% of the heat is lost by the drilling machine. So, the heat left will be,

$Q=\frac{10}{3}\times 0.4$

$Q=\frac{4}{3} J$   (4)

By inserting the values in equation (1) we get;

$\Delta T=\frac{\frac{4}{3} }{10\times 0.504}$

$\Delta T=\frac{4}{3\times 10\times 0.504}$

$\Delta T=0.2646\ \textdegree C$

Hence, the rise in temperature is 0.2646°C.

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