Question

A hot ball of mass 0.2kg is added into 0.5kg of water at 10⁰C. The resulting temperature is 30⁰C. Calculate the temperature of hot ball. Specific heat capacity of iron= 336 J kg-¹ K-¹ and specific heat capacity of water = 4.2 × 10³ J kg-¹ K-¹.
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Answer 1

Solution:-

Here we are given with the values which are written as follows,

• Mass of hot ball = 0.2kg
• Temperature of water at which hot ball is added = 10⁰C
• Resulting temperature = 30⁰C

Let us assume that the temperature of hot ball be t⁰C.

Let us assume that the temperature of hot ball be t⁰C.As we know that there is a fall in temperature so,

==> Temperature fall = ( t - 30⁰ ) C

There is a rise in temperature. So,

==> Temperature rise = (30-10)⁰C

Again given,

• Specific heat capacity of iron= 336 J kg-¹ K-¹
• specific heat capacity of water = 4.2 × 10³ J kg-¹ K-¹.

Here finding out the heat energy possessed by the hot ball:

==> Mass × Specific heat capacity of iron × fall in temperature

Substituting values:

==> 0.2 × 336 × (t-30) J

Now finding out the heat energy undertaken by ball:

==> Mass × Specific heat capacity of water × rise in temperature

Substituting values:

==> 0.5 × (4.2 × 10³) × (30-10)J

At last we are calculating the temperature of hot ball by keeping and solving heat energy given by ball to heat energy taken by water.

==> 0.2 × 336 × (t-30) J = 0.5 × (4.2 × 10³) × (30-10)J

==> 2/10 × 336 × (t-30) = 5/10 × 4/10 × 10³ × 20

==> 67.2t - 2016 = 42000

==> 67.2t = 42000 + 2016

==> 67.2t = 44016

==> t = 44016 / 67.2

==> t = 655⁰

Hence, temperature of hot ball is 655⁰ C

Answer 2

Solution:-

Let temperature of hot ball be f°C

Fall in temperature of ball (f-30)°C

Heat energy given by ball = 0.2 × 336×(t-30)J

Rise in temperature of ball =(30-10)℃

Heat energy taken by water

= 0.5×(4.2×10³) × (30-10)J

$\\ \dashrightarrow \:\bf \red{ Heat\: Energy\: given\: by\: ball\: = \: Heat \: energy\: taken \: by \: water}$

M¹ - C¹ (t¹-t) = M² - C² (t- t²)

$\\ \dashrightarrow \:\bf \blue{ ( 0.2×336)× (t-30) = (0.5×4.2×10³×30)}$

$\\ \dashrightarrow \:\bf \blue{67.2t = 42000+2016 =44,016 }$

$\hookrightarrow \: \bf{t \: = \: \dfrac{44016 \: }{67.2} } = 655℃$

Therefore the temperature of ball is 655℃