Answer:
Detail is given below.
Explanation:
Given data:
Mass of water = 50 g
Initial temperature = 0°C
Final temperature = 100°C
Specific heat capacity of water = 1 cal/g.°C
Calories absorbed = ?
Solution:
Formula:
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 = 100°C - 0°C = 100°C
Now we will put the values in formula.
Q = 50 g×1 cal/g.°C ×100°C
Q = 5000 cal
Hence, it is proved that 5000 calories required to increase the temperature of 50 g of water from 0°C to 100°C.
