Question

What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?

Answer 1

Answer: $74.1^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed=$8.94\times 10^3$ Joules

m= mass of copper = 454 g

c = specific heat capacity = $0.385J/g^0C$

Initial temperature of the copper = $T_i$ = 23.0°C

Final temperature of the water = $T_f$  = ?

Change in temperature ,$\Delta T=T_f-T_i$

Putting in the values, we get:

$8.94\times 10^3=454\times 0.385\times (T_f-23.0)^0C$

$T_f=74.1^0C$

The final temperature of copper will be $74.1^0C$