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

Answer:

T_final = 279.4 [°C]

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]

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