Numa cozinha industrial, a água de um caldeirão é aquecida de 10 ºC a 20 ºC, sendo misturada, em seguida, à água a 80 ºC de um segundo caldeirão, resultando 10 ell de água a 32 ºC, após a mistura. Considere haja troca de calor apenas entre as duas porções de água misturadas e que a densidade absoluta da água, de 1 kg/ ell, não varia com a temperatura, sendo, ainda, seu calor específico c = 1,0,cal,g^{−1} ºC^{−1}. A quantidade de calor recebida pela água do primeiro caldeirão ao ser aquecida até 20 ºC é de

Let's break down the problem step by step:

We have two parts of water, 10 liters at 10°C and 80°C, respectively. After mixing, we get 10 liters of water at 32°C. We need to find the amount of heat received by the water from the first cauldron when it was heated from 10°C to 20°C.

Since the density of water is 1 kg/L, we can calculate the mass of each part of water:

m₁ = 10 L × 1 kg/L = 10 kg (for the 10°C water)

m₂ = 10 L × 1 kg/L = 10 kg (for the 80°C water)

When we mix the two parts, we get:

m = m₁ + m₂ = 10 kg + 10 kg = 20 kg

The temperature of the mixed water is 32°C. We can use the heat capacity formula to find the heat received by the water from the first cauldron:

Q = mcΔT

where Q is the heat received, m is the mass of the water, c is the specific heat capacity of water (1 kcal/kg°C), and ΔT is the temperature change.

For the water from the first cauldron, the temperature change is ΔT = 20°C - 10°C = 10°C.

Plugging in the values, we get:

Q = 10 kg × 1 kcal/kg°C × 10°C = 80 kcal

Therefore, the correct answer is D) 80 kcal.

Explanation: The key to this problem is to recognize that the heat received by the water from the first cauldron is equal to the heat lost by the water from the second cauldron. Since the temperature of the mixed water is 32°C, we can use the heat capacity formula to find the heat received by the water from the first cauldron. The correct answer is 80 kcal.