Final answer:
The amount of heat required to heat 868 g of water from 19°C to 89°C is 48,976.139 calories. The correct answer is d. 48,110 calories.
Step-by-step explanation:
The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. Since 4.184 J is required to heat 1 g of water by 1 °C, we will need 868 times as much to heat 868 g of water by 1 °C. To find the amount of heat required to heat water from one temperature to another, we can use the formula:
q = mcΔT
where:
- q is the amount of heat
- m is the mass of the water
- c is the specific heat of water
- ΔT is the change in temperature
In this case, the change in temperature is 89°C - 19°C = 70 °C. Substituting the given values into the formula:
q = (868 g) x (4.184 J/g °C) x (70 °C) = 204,462.912 J
Converting the result to calories:
204,462.912 J x (1 cal / 4.184 J) = 48,976.139 cal
Therefore, the correct answer is d. 48,110 calories.