3. How much heat does it take to

**Answer 1**

1. 13,500 cal First, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by where m = 150 g is the mass of the ice C_i = 0.5 cal/g C° is the specific heat capacity of the ice is the temperature change of the ice Substituting now we need to find the amount of heat needed to melt the ice, where m = 150 g is the mass of the ice is the latent heat of melting Substitution, so the total heat needed is 2.3750 cal The amount of additional heat needed to heat the water at 25°C is where m = 150 g is the mass of water C_w = 1 cal/g C is the specific heat capacity of water is the change in temperature Substituting, 3. 9200 cal All of First, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As in point 1., this is given by where m = 80 g is the mass of the ice C_i = 0.5 cal/g C° is the specific heat capacity of the ice is the temperature change of the ice Substituting , Now we have to find the amount of heat needed to melt the ice: where m = 80 g is the mass of the ice is the latent heat of fusion By substituting, finally, the amount of heat needed to heat the water to 25 °C is where m = 80 g is the mass of water C_w = 1 cal/g C is the specific heat capacity of water is the change in temperature Substitution, so the total heat required is 4. No Explanation: The total heat required for this process is 3 different amounts of heat: 1- The heat required to bring the ice to melting temperature 2- The heat required to melt the ice while its temperature remains constant 3- The heat required to raise the temperature of the water However, the calculation of the quan tity of heat is necessary to melt the ice and adding the amount of heat necessary to raise the temperature of 80 g of water by 45°C is not equivalent: indeed, the calculation of point 1) requires using the specific heat capacity of ice, not that of water, so the two are not equivalent.

**answer 2**

The latent heat of fusion of water is 334 Joules per gram. Since there are 29.95 grams, we multiply 334 by 29.95 = 10003.3 Joules, which is choice (4) 1.00 * 10^4

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