Question

$0.120$kg of copper of specific heat capacity $100\frac{cal}{K{g}^{o}C}$ and at ${20}^o$C is heated by a burner for two minutes, which supplies heat energy at a rate of $20$ calories/sec. Calculate the final temperature of copper.

Answer 1

m = 0.12kg

s = 100cal/kg/℃

$T_1$ = 20℃

$T_2$ = ?

$△$T = $T_2-T_1$

Heat supplied by burner

$\Rightarrow$ = 20calories/sec

Time heated for = 2 minutes

$\Rightarrow$ = 2$\times$60

$\Rightarrow$ = 120 seconds.

Total heat received

$\Rightarrow$ = Heat per second $\times$ time (in seconds)

$\Rightarrow$ = 120 $\times$20

$\Rightarrow$ = 2400cal.

Q = ms$△$T

2400 = 0.12$\times$100$\times$$△$T

$△$T = $\frac{2400Y}{0.12\times 100}$

$△$T = 200

$T_2-T_1$ = 200

$T_2$-20 = 200

$T_2$ = 200℃.