Question

While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°C), you found that the substance gained 305 calories. What was the final temperature of your sample?
**Round your answer to the nearest 0.1**

I got 83.1 but I just want to check it

Answer 1

Answer:

Final temperature  = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 -  25°C

305 cal = 5.25cal/°C × T2 -  25°C

305 cal / 5.25cal/°C = T2 -  25°C

58.1 °C = T2 -  25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C