Question

To what temperature will a 79.0 g piece of glass raise if it absorbs 9457 calories of heat and its specific heat capacity is 0.50 cal/g°C?? The initial temperature of the glass is 40.0°C. SHOW YOUR WORK *​

Answer 1

T_final = 279.4 [°C]

Step-by-step explanation:

In order to solve this problem, we must use the following equation of thermal energy.

`Q=m*C_(p)*(T_(f)-T_(i))`

where:

Q = heat = 9457 [cal]

m = mass = 79 [g] = 0.079 [kg]

Cp = specific heat = 0.5 [cal/g*°C]

T_initial = initial temperature = 40 [°C]

T_final = final temperature [°C]

`9457 = 79*0.5*(T_(f)-40)\\239.41=T_(f)-40\\\\T_(f)=279.4[C]`