Question

A 500 W heater is totally immersed into a huge block of mass 10 kg with a temperature of 30°C. In 6.3 minutes, the temperature of the metal rises to 80°C. Calculate the specific heat capacity of the metal.

Answer 1

Answer:

$\boxed{\mathfrak{Heat \ capacity \ of \ the \ metal = 378 \ J/kg.K}}$

Explanation:

Mass of substance (m) = 10 kg

Initial temperature ($\rm T_i$) = 30°C = 303 K

Final temperature ($\rm T_f$) = 80°C = 353 K

Power of heater (P) = 500 W

Time (t) = 6.3 min = 378 sec

Heat energy (Q) of heater is given as:

$\boxed{ \bold{Q = mc\Delta T}}$

c → specific heat capacity of the metal

Power (P) is the amount of energy transferred/converted per unit time i.e.

$\rm \implies P = \dfrac{Q}{t} \\ \\ \rm \implies P = \dfrac{mc\Delta T}{t} \\ \\ \rm \implies P = \dfrac{mc(T_i - T_f)}{t}$

By substituting value in the equation we get:

$\rm \implies 500 = \dfrac{10 \times c(353 - 303)}{378} \\ \\ \rm \implies \cancel{500} = \frac{ \cancel{10} \times c \times \cancel{50}}{378} \\ \\ \rm \implies c = 378 \: J/kg.K$

Answer 2

Given :-

• A 500 W heater is totally immersed into a huge block of mass 10 kg with a temperature of 30°C. In 6.3 minutes, the temperature of the metal rises to 80°C.

To Calculate :-

• The specific heat capacity of the metal

Solution :-

We have,

• Mass of substance = 10 kg
• Initial temperature = 303 K
• Final temperature = 353 K
• Power of heater = 500 W
• Time = 6.03 minute = 378 s

$\rm Heat \: energy \: of \: heater \: is \: given \: as = \boxed{\bf Q = mc∆T}$

We know that,

$\boxed{\sf P = \frac{Q}{t}}$

Substitute the value of Q we get,

$:\implies\sf P = \frac{mc∆T}{t}$

$:\implies\boxed{\sf P = \frac{mc(T_i - T_f)}{t}}$

Substitute the values in the equation,

$:\implies\sf500 = \frac{10 \times c(353 - 303)}{378}$

$:\implies\sf\cancel{500} = \frac{\cancel{10} \times c \times\cancel{ 50}}{378}$

$:\implies\sf c =\underline{\boxed{\blue{ \sf378 \: J/kg.K}}}$