Answer:
Therefore the new temperature of the water is 91.67°C
Explanation:
Specific heat of water = 4.186 joule /gram °C
The equation to calculate change of temperature
Q=s×m×ΔT
Q = the amount of heat = 30,000 J
s= Specific heat of water = 4.186 joule /gram °C
m = mass of the water = volume ×density=(100×1)gram =100 gram
ΔT = the change of temperature = (T-20)
Therefore,
30000= 4.186×100×(T-20)
⇔(T-20)= 71.67
⇔T = 71.67+20
⇔T = 91.67
Therefore the new temperature of the water is 91.67°C
