How many grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C?
To find out how many grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C, we need to calculate the amount of heat transferred.
First, we can calculate the heat absorbed by the water at 90°C using the formula:
Q = mcΔT
Where:
Q = heat absorbed (in calories or joules)
m = mass of water (in grams)
c = specific heat capacity of water (1 calorie/gram°C or 4.18 joule/gram°C)
ΔT = change in temperature (in °C)
Assuming the specific heat capacity of water is 1 calorie/gram°C, we can calculate the heat absorbed by the 800g of water at 90°C:
Q1 = (800g) * (1 calorie/gram°C) * (90°C - 50°C)
Q1 = 32000 calories
Next, we need to calculate the heat lost by the water at 20°C when it reaches 50°C:
Q2 = mcΔT
Assuming the specific heat capacity of water is 1 calorie/gram°C, we can calculate the heat lost by the water at 20°C:
Q2 = (m) * (1 calorie/gram°C) * (50°C - 20°C)
Q2 = 30m calories
Since energy is conserved in this process, the heat absorbed by the water at 90°C is equal to the heat lost by the water at 20°C:
Q1 = Q2
32000 calories = 30m calories
Solving for m:
m = (32000 calories) / (30 calories)
m = 1066.67 grams
Therefore, approximately 1066.67 grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C.